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8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 1/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 2/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENT AEY ^/^^^ TEIGONOMETRY BY H. S. HALL, M.A., FOEMERLY SCHOLAR OF CHRIST's COLLEGE, CAMBRIDGE. AND S. R. KNIGHT, B.A., M.B., Ch.B., FORMERLY SCHOLAR OF TRINITY COLLEGE, CAMBRIDGE. FOURTH EDITION, REVISED AND ENLARGED. UonUon CO., MACMILLAN MACMILLAN NEW YORK THE AND : Limited COMPANY 1906 AU niihtif renerved. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 3/467 8/12/2019 Elementary Trigonometry by Hall and Knight First Edition, 1893. Second Edition, Third Edition, 1898, Fonrth Editioti, 1899, revised 1901, and 1S95, 1902, enlarged, 1S96, 1897. 1904. 1905 Reprinted igo6 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 4/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 5/467 8/12/2019 Elementary Trigonometry by Hall and Knight The Tables uf Logarithms and Antilogarithm.s have been taken, with slight inodifications, from those published by the Board of Education, South Kensington. The Four-Figure Tables of Natural and Logarithmic Functions have been For these aui I reduced from Mr indebted to greatly who kindly undertook Seven-Figure Tables. the laborious Frank Castle, a special task of compilation for this book. (9) An easy first teachers to postpone, of identities course has been if they wish, all mapped out enabling but the easier kinds and transformations, so as to reach the more practical parts of the subject as early as possible. All the special features of earlier editions have retained, satisfy all and it is hoped that the present additions Vjeen will modern requirements. H. S. HALL. August 1905. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 6/467 8/12/2019 Elementary Trigonometry by Hall and Knight SUGGESTIONS FOR A FIRST COURSE. In the eighteen iirst jjlaeed before all articles chapters and an been has may examples which sets of conveniently be omitted from a asterisk course. tirst For those who wish to postpone the haixler identities and transformations, so as to reach practical work with Four-Figure Logarithms at an detailed coiirse is I— HI, Chaps. Examples Chaps. IV — eai'lier stage, the following recommended. Arts. 1—30, 32, 33. [Omit Art. 31, in. b.] IX. [Postpone Chaps, xi and xii.] Chaps. XIII— XV, Arts. 137—170, 182^—182^. [Omit Seven-Figure Tables, Arts. 171—182.] Chaps. and XI, XII. [Omit Arts. 127, 136, Examples xi. f. XII. e.] Chap. XVI, Arts. 183—187, tions 197v— 197d. with Seven-Figure Tables, Arts. [Omit Solu188 — 197.] Chaps. XVII, XVIII, Arts. 198—218. From this point the omitted sections must be taken at the discretion of the Teacher. vn http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 7/467 8/12/2019 Elementary Trigonometry by Hall and Knight Digitized by tine Internet Arciiive in 2010 University of witii funding from Britisii Columbia Library http://www.archive.org/details/elementarytrigoOOhall http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 8/467 8/12/2019 Elementary Trigonometry by Hall and Knight CONTENTS. Chapter I. measurement uv angles. Page Definition of Angle 1 ......... Sexagesimal and Centesimal Measures Formula Chapter — = —-. II, . . . . . . . , . 2 . . , . . £ trigonometrical ratios. Definitions of Eatio and Comniensuiable Quantities ... 5 6 Definitions of the Trigonometrical Eatios ..... ..... Sine and cosine are less than unity, secant and cosecant are greater than unity, tangent The trigonometrical and cotangent are unrestricted ratios are independent of the lines which include the angle the lengths of The III. relations between the ratios. reciprocal relations Tangent and cotangent in terms of sine and cosine ... .......... Sine-cosine, tangent-secant, cotangent-cosecant foiinulse Easy Identities Each ratio can be expressed in terms of Chapter IV. any of the others . . . 12 13 14 16 21 trigonometrical ratios of certain angles. ...... ...... ....... Trigonometrical Ratios of 45°, 60°, .S0° Definition of complementary angles The Use 9 10 Definition of function Chapter 7 . of Tables of Natural Functions 24 27 'J .l \ Easy Trigonometrical Equations ?{0 Miscellaneous Examples. 32 A. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 9/467 8/12/2019 Elementary Trigonometry by Hall and Knight CONTENTS. solution of right-angled TRiANOLEa. Chapter V. Case Case I. II. ...... When two sides are given When one side and one acute angle are given 36 . Case of triangle considered as sum or difference of two rightangled triangles ......... Chapter VI. PAGE 38 easy problems. Angles of Elevation and Depression 41 The Mariner's Compass 45 radian or ciroular measure. Chapter VII. Definition of Radian Circumference of circle All radians are equal IT ...... ...... = 27r radians = 2 right angles =: (radius) 49 50 . 51 52 180 degrees Radian contains 57 '2958 degrees 52 - 53 Formula —= Values of the functions of — , 4 - — 3b , Ratios of the complementary angle Radian measure of angles -,. „ —- of a regular Radian measure of an angle = -r^ , 55 . 56 . polygon 56 . subtending arc 58 =-. radius Radian and Circular Measures are equivalent Miscellaneous Examples. . . . 61 . ratios of angles op any srAGNiTUDE. Chapter VIII. ......... Convention of Signs angles B. 58 (1) for line, (2) for Definitions of the trigonometrical ratios of plane surface, any augle (3) for 64 66 . Signs of the trigonometrical ratios in the four quadrants Definition of Coterminal Angles The fundamental of the augle The ambiguity when A is ..... ....... formula' of Chap. . iii. are true for all of sign in cos /I— ^.sJl-sin'Â can be known http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 68 68 ahu 70 removed 71 10/467 8/12/2019 Elementary Trigonometry by Hall and Knight Xl CONTENTS. variations of the functions. Chapter IX. .....••••'* PAGE Detiuition of limit 7* Functions of 0° and 90° Changes in the sign and magnitude of sin .4 Changes in the sign and magnitude of tan A . . . . . . 76 • .78 79 Definition of Circular Functions 79a Graphs of the Functions Miscellaneous Examples. 81 C circular functions of allied angles. Chapter X. -A Circular functions of 180° Definition of supplementary angles + .4 90° + .4 Circular functions of 180° Circular functions of Circular functions of - ...••• 82 83 84 85 ^4 87 Definition of even and odd functions Circular functions of 90° - A for any value of .4 . . .88 89 Circular functions of n. 360° ± J The functions of any angle can be expressed as the tions of same func- 90 some acute angle The number angles of which have the same trigonometrical ^^ ratio is unlimited functions of compound angles. Chapter XI. Expansions of the sine and cosine of A sin (.4 + i?) sin (.4 - U) = sin- .4 - sin2 7? Expansions of tan (.4 Expansions of sin (.4 + B and A-B 98 + B) and cot (.4 + B) +B + C) and tan {A + B + C) . 99 100 Converse use of the Addition Formula Functions of 2.4 102 . 105 Functions oi 3A 106 Value of sin 18° Chapter XII. transformation of products and SUMS Transformation of products into sums or differences Transformation of sums or differences into products Eelations 94 96 when J +5 + = 180° http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight . 110 . 112 118 11/467 8/12/2019 Elementary Trigonometry by Hall and Knight CONTENTS. Xll relations between the sides and angles Chapter XIII. OF A TRIANGLE. _ b _ C A~ sin B~ sin (I sin a^:=b^-i-c- a = b cos C -2bc cos A, and C+c The above ~ — cos.-l 2bc B cos When I. Case II. Case III. .... .... ..... them When When two angles and a side are given two sides and an angle Sometimes are given. ojjposite this case is The Ambiguous Case discussed by Chapter XIV. A'^ lirst to 127 . 128 one of .... ambiguous . . finding the third side 131 . are equivalent characteristic of a common logarithm may .... be written 139 140 down by inspection digits logjJV= — 142 of all numbers which have the have the same mantissa ^ 135 139 of a product, quotient, jJower, root The logarithms 1.30 138 a'°^'' '^=A^ is identically true The . . . logarithms. and x = log,. iV^ Logarithm 126 D Miscellaneous Examples. = 125 126 The Ambiguous Case discussed geometrically a'' 124 the three sides are given Wlien two sides and the included angle are given Case IV. 124 independent sets of formula; are not Solution of Triangles without logarithms Case 123 xlogaJ^, and logj a x log„ & = 1 same significant 143 .... 146 E.xponential Ecjuations 148 E Miscellaneous Examples. Chapter XV. the of iakjarithmic tables. u.sk Eule of Proportional Parts Common Use of Tables of Use of Tables of Natural 150 , ....... ..... LogarithniK and Logaritlimic Functions Use of Four-Figure Tables http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight . . 152 152 156 163^ 12/467 8/12/2019 Elementary Trigonometry by Hall and Knight Xm OONTKNTS. solution of triangles with logarithms. Chapter XVI. PAGK Functions of the half-angles in terms of the sides Sin A . . 164 . 166 in terms of the sides Solution Solution when the three sides are given when two sides and the included angle Solution when two angles and a Solution when two sides side are given 167 170 .... are given and the angle opposite to . . 174 one of them 175 are given Adaptation of Adaptation of + ^- to logarithmic work a^ + h- - 2ab cos C to logarithmic work 177 rt2 Solution of triangles with Four-Figure logarithms . . . . 178 . . 183^ heights and distances. Chapter XVII. ...... ..... ..... .184 Measurements in one plane Problems dependent on Geometry 189 Measurements in more than one plane 193 Problems requiring Four-Figure Tables 197 Chapter XVIII. properties op triangles and polygons. Area of a triangle Eadius of the circum-circle of a triangle Radius of the ..... in-circle of a triangle relations established 202 by Geometry . . . Inscribed and circumscribed polygons Area of a circle and sector of a circle The Ex-central Triangle The Pedal Triangle ...... 210 212 .... 216 Diagonals and circura-radius of a cyclic quadrilateral Chapter XIX. . , ........ Distance of ortliocentre from circum-centre Miscellaneous Examples. 204 208 Distances of iu-centre and ex-centres from circum-centre Area of any quadrilateral 200 201 Eadii of the ex-cLrcles of a triangle Some important 198 . . F 214 218 220 222 228 general values and inverse functioks. .... Fonnula for all angles which have a given sine Formula for all angles which have a given cosine http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight . . 232 233 13/467 8/12/2019 Elementary Trigonometry by Hall and Knight XIV CONTENTS. PAGE Formula for all angles Formula for angles which have a given tangent both equi-sinal and equi-cosinal . . . 234 . . . 234 General solution of equations 236 Inverse Circiilar Functions 238 Solution of equations expressed in inverse notation Miscellaneous Examples. G Chapter XX. . . . functions of submultiple angles. Trigonometrical Eatioa of - 247 8 Given cos To A and sin— to find yl express sin A — A — and cos in cos A — 248 terms of sin Variation in sign and magnitude of cos Sine and cosine of 9° To find tan — when A 9< ~ When ^ If cos , sin 6, 0, A is given -— 2 , and sin . . . 250 . . . . 254 ...... — . -cos- cos g 256 258 arc in ascending order of magnitude 0>0- .... — 201 262 265 4 266 cos sin^ -^ --g -—— decreases from O . 254 and approximations. Value of sin 10 cos . . 259 = 0, -—— = 1, and ——- =1 0>1- . — limits tan sin to find the functions of to find cos Chapter XXI. - A . tan Given a function of A Given cos 244 246 1 to — . . as ^ increases from http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight . to — 2 IT Distance and Dip of the Visible Horizon . . 266 267 269 14/467 8/12/2019 Elementary Trigonometry by Hall and Knight CONTENTS. Chapter XXII. Expansiou of Uin Formulfe (tEometrical proofs. (.^ t PAGE 273 7?) for transformation of 274 into i>roiliu-ts sinii'f 276 Proof of the 2A formula' Value of sin 18° 277 Proofs by Projection 278 ...... 282 General analytical proof of the Addition Formula^ Miscellaneous Examples. Graphs of sin 6, tan 6, see If «r='',-fi-'V. then.*? of the sines 283 285 . summation op finite series. Chapter XXIII. Sum H = .;„+ ....... -7'i and cosines of a series of v angles in a. r. 288 289 . 0 -_ AVlien the Sum common difference of the squares and cubes series of ancles in a. r. Chapter XXIV. is , n the sum of the sines . . . is zero . and cosines . . of • 290 . a 293 • miscellaneous transformations and identities. Symmetrical Expressions. Alternating Expressions Allied formulae in Algebra Identities derived Chapter XXV. InequaUties. .... ........ .... S and 11 notation and Trigonometry 296 303 306 308 by substitution miscellaneous THEORE>rs and examples. 313 Maxima and Minima 319 Elimination Application of Trigonometry to Theory of Equations Application of Theory of Equations to Trigonometiy Miscellaneous Examples. . . 32G . . 328 33G T K. • 337 Miscellaneous Examples. Tables of Functions Answkrs ...... ......... Logarithms, Antilogarithms, Natural and Logarithmic .... http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight .... 374 391 15/467 8/12/2019 Elementary Trigonometry by Hall and Knight c C http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 16/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTAKY TRIGONOMETKY, I. CHAPTER MEASUREMENT OF ANGLES. The word Trigonouietiy in its primary sense signifies the measiu-enaent of triangles. From an early date the science 1. also included the establishment of the relations which subsist between the sides, angles, and area of a triangle but now it has embraces all geometrical wider of scope a and manner much algebraical investigations carried on through the mediumand of ; which will be defined in Chap. II. In every branch of Higher Mathematics, whether Pure or Ajjplied, a knowledge of Trigonometry is of the certain quantities called trigonometrical ratios, greatest value. Suppose that the straight line 01' Definition of Angle. in the figure is capable of revolving about the point 0, and 2. suppose that in this way it has passed successively from the position OA to the positions occupied by OB, OC, OD, ..., then the angle between OA and any position such as OG is measured by the amount of revolution which the line OP has undergone in pa^^sing fi'om its initial position OA into its final position OC. Moreover the line OP may make any number of complete revolutions through the original position Oxi before taking vqy its final position. H. K. E. T. ^ http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 17/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 18/467 8/12/2019 Elementary Trigonometry by Hall and Knight MEASUREMENT OF ANGLES. 1.] 3 work it is conveiiieut to use another method of uieasiirement, where the unit is the angle subtended at the lu theoretical centre of a circle an arc whose length Vjv equal to the radius. is This system is known as Circular or Eadian Measure, and be fully explained in Chapter VII. An angle is will usually represented by a single letter, different letters A, B, C,..., a,^, y,..., 6, being used to distinguish different angles. For angles estimated in sexagesimal or centesimal measure these letters are used indifferent)}', but we shall always denote angles in circular measure by letters taken from the Greek alphabet. If the u umber of degrees and grades contained in an angle 7. be yj/,..., cf), D and G T) respectively, to prove that -^ C = tt. • when In sexagesimal measui'e, the given angle is measure, the same fraction denoted by is -^ ..-^ —- In centesimal . ; 100 ; thatis,- To pass from one system 8. fu'st — — denoted by the fraction of a right angle expi-essed as to = -^^. the other it is advisable to express the given angle in terms of a right angle. In centesimal measure any number of grades, minutes, and seconds may be immediately expressed as the decimal of a right Thus angle. 23 grades 15 minutes angle = ^^ = jj^Q^g of a right angle of a grade = *15 = '23 of a right angle of a grade = '0015 ; of a right ; .•. Similarly, 15^ 23^ 15' -= '2315 of a right angle. 7' 53*4 = 1507534 of a right angle. Conversely, any decimal of a right angle can pressed in grades, minutes, and seconds. Ije at nnce ex- Thus •2173025 of a right angle = 21 •73025« = 2P 73-025' -21« 73 2-5 . In [>ractice the intermediate steps are omitted. 1—2 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 19/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 4 Example Keduce 2^13' 45 1. I. sexagesimal measure. -to This angle = -0213045 of a right angle = 1° 55' 2-658 [CHAP. of -0213045 . 1-917405 • a right angle degrees 60 56-0443 minutes 60 2-658 seconds. In the Answers we shall express the angles to the nearest tenth of a second, so that the above result would be written 1° 55' 2-7 . Obs. Example Eeduce 12° 2. 13' 14-3 to centesimal measure. Thisangle = -13578487...ofarightangle _ 13^57 = 60 60 90 84-9. 1^-3 seconds ) 13-238333.. niiuutes ) . 12-22063b8 .. degrees ) •13578487 EXAMPLES. . . .of a riglit angle. I. Express as the decimal of a right angle 1. 67° 4. 2° 10' 12 . 30'. 2. 11° 15'. 5. 8° 0' 36 . 3. Keduce to centesimal measure 7. 69° 13' 30 . 8. 19° 0' 45 . 10. 43° 52' 38-1 . 11. 11° 0' 38-4 . 13. 12' 9 . 14. 3' 26-3 . Reduce to sexagesimal measure 15. 56^87^50 . 16. 395 6^25 . 17. 40M' 18. 1^ 2' 3 . 19. 3^ 2* b'\ 20. 8^ 10' 6-5 . 21. 6' 22. 37' 5 . 23. The sum 25^\ of two angles is and their 80^ 25-4 . difference is 18°; find the angles in degrees. The number of degrees in a number of grades in the angle is 152 24. 25. and in 26. certain angle added to the : what is the angle 1 same angle contains in English measiu-e x minutes, French measure y minutes, prove that bOx=21y. If the If s and t respectively denote the numbers of sexagesimal and centesimal seconds in any angle, prove that 250« = 81i;. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 20/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight CHAPTER II. TRIGONOMETRICAL RATIOS. Definition. Ratio is the relation which one quantity bears to another of the sa7ne kind, the comparison being made by considering what multiple, part or parts, one quantity is of the other. 9. To by find ; what multiple or part hence the ratio of fraction of is 5 we divide A to A B ^ be measiu'ed by the B may p. In order to compare two quantities they must be expressed in terms of the same unit. Thus the ratio of 2 yards to 2 X 3 X 12 8 27 inches Obs. is — measured by the fraction —; 27 •^ Since a ratio expresses the quantity contains another, every ratio is or 3 number of times that one a numerical qua7ititij. Definition. If the ratio of any two quantities can be exactly the ratio of two integers the quantities are by expressed otherwise, they are said to be said to be commensurable incommensurable. For instance, the quantities 8i and 5\ are commensurable, while the quantities y/2 and 3 are incommensurable. But by finding the numerical value of ^/2 we may 10. ; express the value of the ratio ^'2 3 by the ratio of two commensurable quantities to any required degree of approximation. Thus to 5 decimal places ^^ = 1 '41421, and therefore to the : .same degree of approximation J2 : 3 = 1-41421 Similarly, for the ratio of : 3 = 141421 : 300000. any two incommensural)lo http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight quantities. 21/467 8/12/2019 Elementary Trigonometry by Hall and Knight la.KMKNTARV TRIGONOMKTH V. [I'HAP. Trigonometrical Ratios. Let 11. angle ary ; JÂQ be any acute JP one of the boundtake a point B and perpendicular to AQ. in lines draw BC Thus a right-angled is triangle BAC formed. With reference to the angle A definitions the following are ern- ]iloyed. The ratio BC 1 ^. riie ratio „,, , . T lie ratio ,. ,„, 'I lie I'atio mi The i- ratio opposite side is called tlie sine of B AC -^ AB hypotenuse udnacent -~ or hypotenuse BC — opposite side -—^ or m „ . , adjacent side AC y^^ n( side 7or adjacent ryopposite side — is c;illed —r^ adjacent is called A hypotenuse ^f- i.r ( AB . opposite side ratios are . ,, . , ,, , ,, , side hypotenuse '^^ r^ BC _ . .,,.-,. * the tangent . AC -^^ -- ^, called the cosine of rj- is called ' y-. „, ^. Tlie latio ^jy. or 'J'hese six side A. . is . A. * a 01 A. ,, 4. ^ the cotangent of . a A, e the secant of A. ,, ,, k ». ciJled the ^ c k COSecant 01 A. as the known as long that the angle remains the astrigonometrical be shewn later same the trigonometrical ratios remain the same. [Art. 19.] It will 12. In.stead of writing in full the ratios. words sine, cosine, tangent, Thus the conveniently expressed and cotangent, secant, cosecant, abljreviations are adopted. above definitions may be arranged as follows more : , - in A cosec http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight AB A 22/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight TRTdONOMRTRTOAT, RATIOS. jxT i Tu addition to these six ratios, two others, the versed sine and coversed sine are sometimes used they are written vers A ; and covers A and are thus defined vers ^^^^^ .4 = 1- cos A , covers .4 = 1- si n A In Chapter VIII. the definitions of the trigonometrical any magnitude, ratios' will be extended to the case of angles of but for the present we confine our attention to the consideration 13. of acute angles. 14. Although the verbal form of the definitions of the trigonometrical ratios given in Art. 11 may be helpful to the will gain no freedom in their use until he is first, student at he able to write down from the figure any ratio at sight. In the adjoining PQR is a which P$ = 13, figure, right-angled triangle in the greatest side, R is The trigonometrical the right angle. and Q may be ratios of the angles Since PQ is F at once written ; for example, down „ QR „ 12 important to observe that of an angle are numerical quantities. 15. It is PQ 13 the trigonometrical ratios Each one of them presents the ratio of one length to another, selves never be regarded as lengths. re- and they must them- In every right-angled triangle the hypotenuse is the greatest side; hence from the definitions of Art. 11 it will be seen that those ratios which have the hypotenuse in the denominato-r can never be greater than unity, while those which have the hypotenuse in the numerator can never be less than 16. unity. Those ratios which do not involve the hypotenuse are not thus restricted in value, for either of the two sides which Hence subtend the acute angles may be the greater. the sine and the cosecant the tangent cosine of and an angle can never secant of he greater than an angle can never he less than and cotangent may have any numerical http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 1 1 ; ; value. 23/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 24/467 8/12/2019 Elementary Trigonometry by Hall and Knight TRIGONOMKTRIOAL RATIOS TI.] EXAMPLES 7j^ 11. a. [App7'0.vimate results should be given to two places of decimals.^ Draw an 1. •A'alue of and and find by measurement the cosine. Construct an angle of 39°, and find the value of 2. /ând sine its angle of 77°, its sine cosine. The 3. ./value of sine of an angle is '88 ; draw the angle and find the its cosine. Construct an angle who.se cosine is •34; measure the angle to the nearest degree, and find its sine and tangent. 4. 5. Draw an 6. Given sec^4 angle of 42°, and find = 2'8, its tangent and shie. draw the angle and measure it to the nearest degree. Construct an angle whose sine to the nearest degree. 7. , .' i;^. 6; measure the angle Consti'uct an angle from each of the following data 8. (ii) (iii) tan .4 = -7; cosi?=-9; sinC=-71. In each case measure the angle to the nearest degree. (i) Find sinyl, tan 5, cos C. Construct an angle ^ such that tan .(4 = 1 '6. Measure the angle to the nearest degree, and find its sine and cosine. 9. Construct a triangle ABC, right-angled at C, having the Measure A C and hypotenuse 10 cm. in length, and tan A = 'SI. and find the values of sin A and cos A. the angle 10. ; 11. is 34. sin .1 A Find the cosine and cosecant of an angle A whose sine Prove that the values appro.ximately satisfy the relation cosec 12. .4 = 1. Draw L A CB = 72°. having BC=H cm., lABC=5S\ Draw and measure the altitude, and hence find a triangle ABC approximately the values of tan 13. Draw 53°, cot 72°. a right-angled triangle .4 data tan.4 Measure c = -7, A(?=90°, ^ = 2-8 from the following cm. and the lA. Draw the angles who.sc sines are -Cu and Measure thoir difference of a common arm. 14. side /](' http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 94 on the same in degree.s. 25/467 ' 8/12/2019 Elementary Trigonometry by Hall and Knight KIiEMENTARV TRIGONOMETRY. Let 17. angle at A ; the sq. on ABC be a right-angled then by Geometry, tri.ingle [chap. having tlie riglit BC of or, = sum more sqq. on and A AC briefly, /?, BC^ = AC^ + AB-\ use this latter mode of expression it is understood that the sides AB, AC, BC are expressed in When we terms of some common unit, and the above statement may be regarded as a numerical relation connecting the numbers of units of length in the three sides of a right-angled triangle. numbers of units of length in the sides opposite the angles A, B, G by the letters a, b, c respectively. Thus in the above figure we have rt- = ?/^-|-c2, so that if the lengths of two sides of a right-angled triangle are known, this equation It is usual to denote the will give the length of the third side. Exaviple ABC 1. which C find c, and also of • is Here c -ô^- Also h-^ if a — 3, h = cof,B = — 4, cot-B. = {Sf+{Af = 9+U = 2r, A sin a right-angled triangle J and sin + is right angle; the BC AB BC ; 3 5' 3 1 AC A ladder 17 ft. long is placed with its foot at a Exaviple 2. distance of 8 ft. from the wall of a house and just reaches a windowFind the height of the wiudow-sill, and the sine and tangent sill. of the angle which the ladder makes with the wall. Let ^C be the ladder, and Let X be the number of then X- = [llf - (8)2= (17 .-. Also sin a; C BC feet in + 8) the wall. BC\ (17 - 8) =2 , x 9 = 5x3 = 15. = AB AC 8 17' 8^ i ') http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 26/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 27/467 8/12/2019 Elementary Trigonometry by Hall and Knight KLEMENTARY TRIGONOMETRY. 10 [CHAP. A denote any acute angle, we have proved that all the trigonometrical ratios of A depend only on the magnitude of the angle A and not upon the lengths of the lines which bound It may easily be seen that a change made in the the angle. 20. If value of A will produce a consequent change in the values of all This point will be discussed the trigonometrical ratios of A. more Chap. IX. fully in Any expression which involves a variable Definition. quantity x, and whose value is dependent on that of x is called a function of x. Hence the trigonometrical trigonometrical functions ; ratios may for the present also we be defined as shall chiefly em- ploy the term ratio, but in a later part of the subject the idea of ratio is gradually lost and the term function becomes more appropriate. The use proved in Art. 19 is well shewn in the following example, where the trigonometrical ratios are employed as a connecting link between the lines and angles. 21. A BG Example. angle. D: \i BD is of the principle is drawn perpendicular AB = 1% JC=16, BC = 2Q, From the angle CBD, right-angled BD BC = tan A is the right BC and meets CA produced in a right-angled triangle of which to find iiD and CD. tri- G] from the http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 28/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight EXAMPLES. The 1. sides AB, BG, CA 15, 8 respectively; write sec 5. The 2. 12 5, U TRIGONOMETRICAL RATIOS. 11.] sides the values of sin J, sec J, tanZf, RP of a right-angled triangle are 13, down the write : of a right-angled triangle are 17, down FQ, QR, respectively II. b. values of cot P, cosec Q, cos Q, cos P. ABC 3. c = 20, is a triangle in which 4. is a triangle ABC = 6 25, find c, sin C, tan J, The 5. 37, 1 A is a right angle; if 6=15, is a right angle; if a= find a, sin C, cos B, cot C, sec C. sides ED, EF, 2 respectively : write in which cosec A. DF of 5 24, a right-angled triangle are 35, values of sec E, sec F, cot E, down the sin/'. The hypotenuse 6. of a right-angled triangle is 15 inches, and one of the sides is 9 inches find the third side and the cosine and tangent of the angle opposite to it. : sine, Find the hjôteiuise AB oi a right-angled triangle in which AC—1, BC=24:. Write down the sine and cosine of A, and shew that the sum of their squares is equal to 1 7. A 8. ladder 41 ft. long is placed with its foot at a distance of from the wall of a house and just reaches a window-sill. Find the height of the window-sill, and the sine and cotangent of the angle which the ladder makes with the ground. 9 ft. A ladder is 29 ft. long how far must its foot be placed from a wall so that the ladder may just reach the top of the wall which is 21 ft. from the ground 1 Write down all the trigono9. ; metrical ratios of the angle between the ladder and the wall. 10. AD : ABCD find all 11. a square C is joined to E, the middle point of the trigonometrical ratios of the angle ECD. ABCD is is ; a quadrilateral in which the diagonal right angles to each of the sides AB, CD: if JZ?=15, = Bb, find sin ABC, sec ACB, cos CD A, cosec AD DAC 12. PQRS right angle. is a quadrilateral in which the angle If the diagonal RP =20, RQ = 2\, RS=16, cosec PR find is sin at JC=36, PSR at right angles to PR>% tanRPS, ^C is RQ, con is a atid RPQ, PQR. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 29/467 ' 8/12/2019 . Elementary Trigonometry by Hall and Knight / CHAPTER III. RELATIONS BETWEEN THE TRIGONOMETRICAL RATIOS. 22. Reciprocal relations between certain ratios. (1) Let ABCh^ and cosec • . .Sin . BC , . then A .4 = a BC~ a' X cosec ^4 =- x a c Thus sin A and . and (2) A cosec I A -= cosec A— A cosec ; ' 1 sin A Again, cos A = AC b —= = -, AB c cos COS iy,) — are reciprocals sin . C a triangle, right-angled at AA = A i . and &ecA= X sec . ^ sec BC AC .• . ( and 1 , A — -7-^ — , A =~^ X c A , Aviso tan , tan and , tan ^4 A X cot ^4 = y — —,-. cot =l JJ ^l b .'. f-, 1 ; 1 -^ , cos A AC ... = — cot a -,- sec ^ Y AB c -ry,== AC A , -„^/^ BC x - a — 1 and cotJ=: http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight h - a ; ; —j tsuiA . 30/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight HKLATIONS BKTWKKN THK TRIUUNOMETRICAL RATIOS. From BC tan.l Again, cot ^4 A cos A sin = , Prove Example. aiul cos A. b ^ c ; '; BC a c . .1 — rj sm J from the tliat cosec A AC^h^h also evident is in terni^ of sin I = cos J -^siIi cos A ^ cotJ= . . a a cc = -j-^=^ = = sin A -r cos A A= tan A cot we have the adjoining figure , which A and To express tan 23. 13 cosec relation ct)t reciin'ocal ^ = , tan A , A tan A = sec A 1 A tan A .1 A _ cos .4 A sin 1 sin A cos -sec A. We frequently meet with expressions which involve the square and other powers of the trigonometrical ratios, such as It is usual to write these in the shorter (sinJ[)2, (tan.4)^... 24. forms sin^^, tan^^l, ... '^ /sin A\ -j tau2 A = (tan Af = \cos A Thus ^ (sin { ^ J Ay __ sin'Â ' (gosAY Shew Example. cos^^4 A that sin- .4 sec sm-^4 sec J . cot-.4 , cot- .4 =sin-.4 x —cos A. 1 x -sin-^ X -COS by cancelling factor.s common to cos t | cos- .4 1 ., -; \s\vi.Aj cos^ . /cos.4\I T ^4 X ., sin-. , .(, numerator http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight anil (leiioininator. 31/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 14 To prove that 25. BA C Let A + cos^ A = \. / be any acute angle AC, and denote the a, b, sin^ [chap. BC draw ; perpendicular to sides of the right-angled triangle ABC by c. By •^ detinition,' and . . . ., , sni- î sin A = -j-^ = cos ^1 - AB c AC _b^ AWc • + cos- ^1 = , - « -„- c- ; ' b+—= + b^ a- , 5 c- cr = 1. = 1- cos ^ A, 008^^4 = 1 -sin^^l, sin- ^1 Cor. Exa7nple 1. = (cos-^ = cos^^ first Example 2. ^ 1 cos J = \/l ; -sin^^. Prove that cos^^J - sin'*^ ^cos^^ - sin^ J. cos*^ - sin*^ since the = V 1 — cos^ A sin + sin2^)(cos- J - - sin^.^) sin^.4, factor is equal to 1. Prove that cot a - cos- a cot a ,^1 J\ - cos-a = cot = = cos a. a x sin a cos a — -; http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight X sin a = cos a. 32/467 , 8/12/2019 ., . . Elementary Trigonometry by Hall and Knight RKLATIOXS BF,TWf:KN THK TRIGONOMETRICAL RATIOS. III.] 26. To pt-ove that With the sec^ / A = l+tan^ A. we have figure of the previous article, sec A C = Yb A = -J--, c2 , ., IT) ; b'^ + a^ -l+taii',4. ( A - tan- .4 = 1, tan^ J^ = sec2il - 1, .sec- 'or. K.fami)le. Prove that cos A co.s ,,^/sec -' A sec tan ^Jsec^ ^ - 1 A = \/l+tan''yl = \/sec2 /I - J. 1 A-l = sin A = cos A x tan A sin ^ COB A7 , = C0S-4 sin A, — 27 . AVîth — \-\- c(jt- J To prove that cosec^ A the figure of Art. 25, cosec X / . we have J = -fr^ = - BC a a'' ; a'' a- = r'< iR. cot- .4 = A = cosec^ .4 - ccseccot2 Exnmplf. J - l+cot2.-l. 1, 1 (-oseo A = \/l+cof-.4, cot ^ = ^/cosec''* ^ — Prove that cot> a ~ 1 — cosec* a - 2 cosec^ cot* a - 1 = (cot- a + 1) (cot- a - 1) = cosec^ a (cosec- a - 1 - 1 a. 1) = cosec^ o (cosec- a -2) — cosec'' a - 2 cosec- a. H. K. K. T. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 33/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 16 28. The formulae proved in the last three [CHAP, articles are not independent, for they are merely different ways of expressing in trigonometrical symbols the property of a right-angled triangle known 29. as the Theorem It will of Pythagoras. be useful here to collect the formulae proved in this chapter. T. (;osec^4 X sin J = 1, coseCj4 = -7 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight — r, sin ^1 sin^l: cosec .i 34/467 . 8/12/2019 . Elementary Trigonometry by Hall and Knight KASY IDENTITIES. III.] Example The form 17 - tan- d + tan** 6. Prove that sec* 6 - sec^ 6 2. ~ of this identity at once suggests that we should use the secant-tangent formula of Art. 26; hence the side first = (sec* ^ sec- - 1) = (1 + tan2 6) tan^ e = tan2^ + tan^^. EXAMPLES. Prove the follDwing identities ^ , : 1. sin i4 cot J = cos ^4. 3, cot A sec A =cosec 5. cos yl cosec 7. (1 8. (l-sin2yl)sec2.4=l. 9. cot2^(l-cos2<9) = cos2^. y ^-1 - cos2 ^) = cot )< y A. i4 V . J = cosec^ III. a. 1 . : tan 4. sin A sec .4 =tan A. 6. cot /I sec .4 sin cosec a 13. ( 15. (l-cos2<9)(l-f-tan2^)=:tan-^. 16. cos a cosec a 17. si n2 yl ( 1 19. (l-cos2.4)(l+cot-^l) = 20. sin a sec a Vcosec^ a 21. cos a \/cot2 a 22. Rin2 e cot2 e 23. (l-|-tan2^)(l-sin2^) 24. sin2^sec2^ + tan2 25. (^cosec2 <9 . 1 ) cos2 = sin a. J= ^ J =1. / X — (sec^ 14. 1 1 + cot2 A) = ^. 1 )<^ = cot d. V sec^ a — -I- .-1. \ 12. I /I / tan a '^l-sinâ a = sin /I 11. — sin^ 4>S^ cos (l-cos2^)sec2(9 = tan2(9. ^ 1 ^ 2. 10. V ^ = ^ 1 ) /( - cot^ ^l -1 . /, K'^ 1 X' 1 .1 (cosec^ 18. l. = 1. = Vcosec^ o — 1 ) tanS A-]. ^ ''^ 1 . ^ + sin2 ^ = 1. = l. = sec2^-l. tan2 ^ - 1 ^ tan2 ,< 2 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 2 35/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 36/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight EASY IIJENTITIKS. III.J ,01,1 rovo 7. Kxiuttplc ,,,, ihe .. ii . , — ni'«t side taua-cot/3 , — tau a tan /3 - cot a X tliiit .i. tan a cot - 1 _ ti , — tan a ^ tan a - cot li 1 X 1 — tan a cot The transformations tau a - cot , , cot /:i. li . tan o - cot fi tan a cot ft 1 cot 19 tan a cot B -- tan a - cot /S ft. in the successive stejjs are usually suggested by the form into which we wish to bring the result. For instance, in this last example we might have proved the identity by substituting and cotangent in terms of the sine and cosine. This not the best method, for the form in which the right-hand side is given suggests that we should retain tan a and cot ft unchanged throughout the work. for the tangent however is *EXAMPLES. Prove the following identities siu , 1. \ a a cos a — vers d .sec^ = tan a : = sin d cot — vers d sec 6 5. sec - tau 6. tan 3 + cot = sec 6 cosec 6. 7. VlTcotM VsecM-l \/l-suiM = 8. (cos sin 6 - 1. = cos 6. . + (cot sin2 A (1 12. cos^ A (sec^ 13. cot^ a 15. = sec 6 — J + sin ^)2 + (cos - sin 0)- = 2. (l+taa^)- + (l-tau^)2 = 2sec^<9. 11. . 4. . (cot ^ 14. ^. , a = tana. a cot s cosec' a „ 2. . 1 10. . : 3. 9. f cot^ IILb. 1 )^ (9 + +cot2 A) + cos^ ^ A- cot* a + tan^ A a 1 -, , . + cot^ a r5 l+tan''a cot- , ) -f cosec2 (1 sin^ ^ . sill ,, 0. + tan2 A) = 2. A = cosec* a — cosec'^ —= — + l+sma —= l-sin« tan''^ vv-;^ =2 1 )2 (cosec^ A - cot^ J = ) 1. a. ., « 2 sec2 sec a. a. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 37/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 38/467 ' . 8/12/2019 Elementary Trigonometry by Hall and Knight RELATIONS BETWEEN THE TRIGONOMETRICAL RATIOS. III.] of the relations collected together in Ai't. 29, trigonometrical ratios can be expressed in terms of any- By means 32. all 21 the one. Example of tan Express 1. A We have cot A — tan^ -. all the trigouometrical ratios of cos^ . sin .1 = , A sec — sinA r cos A— cosec = ; =^» , ^14- tan2 A cos ^ A 10,11 â A = tan . . Jt cos CUB sin A ^^^-^ A— , xl , ^/1 .4 1 iu ; A — Jl + tan^ A sec terms A + tan -^ Jl + tanÂ tan A In writing down the ratios we choose the simplest and most reciprocal natural order. For instance, cot A is obtained at once by the imrnecomes sec^ cotangent: tangent and relation connecting the remaining three ratios diately from the tangent-secant formula the Obs. ; now readily follow. Example cosec A— Given cos A 2. 5 — ^^, find cosec J and cot^. 1 sin A Ji- cosÂ 1 'iil cot A — -.—-. — cos A suxA 5 33. Jt. is 13 13 1 1 1 _ ~ 12 _ ~ 12 X cosec A 5 always possible token two sides are given: for a riy lit -angled triangle the third side can be found by to describe (ieometry, and the triangle can then be constructed practically. We can thus r&idily obtain all the trigonometrical ratios when one is given, or expics,s all in terms of any one. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 39/467 , 8/12/2019 . Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGOKOMETRY. 22 E J ample Take a right-angled the riixht angle, and PQ — 5 Let Q/i A=— Given cos 1. triangle , find cosec PQK, having the hypotenuse A and of which PR — 13 Q [chap. cot is units, units. = units .c then ; -(5)2= (lb +5) (13 -5) .1-2^ (13)2 = 18x8 = 9x2x8; = 3x4 = 12. .-. Now a; liPQ = cos PR~13' IRPQ-A. so that . and cosec but A—c A cosec '• QR = PQ PQ J by and A Art. 32, Ex. 2.] in terms of cosec A c. V PR QR' QR-1- he taken as the unit of measurement; l, and therefore PJ? = c contain x units; then x^^c^ - Hence [Compare P=12- i\ienQR = Let 5 PQR right-angled For / RPQ = A. shortness, denote cosec Then QR~ 12' Find tan A and cos 2. Take a triangle Q, and having Let . '=°*^ Example at PR_'^ cosec A Hence tan^: cos.l QR_ 1, so that X 1 PQ ~ Vc^- = -p- = = Jc^ - 1. 1 i ~ c^-1 _ VcoeecM-1 i^cosec -' cosec http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight A-i A 40/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight RELATIONS BETWEEN THE TRIGONOMETRICAL RATIOS. III.] EXAMPLES. 1. (}iven .siii.l— 2. (Jiveii 3. Find cot^ and — J tail tana = -, sec tiiid , Hnd , t>hi tind sin .1 ,1 III. c. and cot.l. and 23 cu.s.l. when .seo^^4. 6 a and cosuu 4. Jf 5. Find the sine and cotaiigeut of an angle whose 6. If 25 sin 7. Express sin 8. Express cosec a and cos a in terms of cot 9. Find sin d and cot 6 .sec <«. .secant is 7. 10. Express ^.4 = 7, find tan A and tan in J and ^1 in sec^^. terms of terms of sec a. d. the trigonometrical ratios of all A co.s .1 terms of in sin A. - cos >1 11. Given sin 12. If sin 13. If 14. r When secJ = —2m ^1 = 0, find cosec J. V J=— n pcot (9 , prove that \/«^ = V$''^-io find sin ^ — >'i^ • tû Î = '• (9. find tan ^1 and 15. Given tan.l=~„^-„, find cos .1 and cosec J. 16. If seca IT ^ a le If cot 17. , P'-r 13 = -T-, o P 6=q , , 2 sin sin J. a- 3 cos a — 4 sin a — 9 cos a . , find the vahie of - i COS ^ JO a ^ ^.x, i find the value of ^- pcos 6+ http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight ^ . —sin ^ $- ~ q -. ;,. — sm d 41/467 8/12/2019 Elementary Trigonometry by Hall and Knight CHAPTER IV. TRIGONOMETRICAL RATIOS OF CERTAIN ANGLES. 34. Trigonometrical Ratios of 45°. right-angled isosceles triangle, with the right be a angle atBAG C; so that Let B=A = 45°. Let each of the equal sides contain I AC=BC=l. then Also AB-' = l' + P = 2P; .: Sin 45 COS 45 AB = l^2. =jg = ^-^ A The other l- i three ratios are the reciprocals of these cosec 45° may J-2' =2^ = ^2 = 72' tan 45 =-77s=7 = or they iruits, be I'ead — J2, oft sec 45° from the = ^^2, cot 45° ; thus =1 figure. r http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 42/467 8/12/2019 Elementary Trigonometry by Hall and Knight TRIGONOMETRICAL RATIOS OF CERTAIN ANGLES. Trigonometrical Ratios of 60° and 35. Let ABC be an equilateral triaugle ; 25 30°. thus each of its augles is 60°. ' Bisect L By spects Eiic. ; BAG by AD I. 4, therefore meeting the triangles BD = CD^ BG at D ; ABD, AGD if then i BAD='i(f. are equal in all re- and the angles at D are right angles. In the http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 43/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 44/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight COMPLBMRN'I'ARY ANOLRK. IV.] 5. 60°. 2 sin 30° cos 30° cot 6. 60 . tan2 45° sin 60° tan 30° tan^ 7. tan2 60° 8. + 4 cos^ 45° -f 3 sec^ 30°. 60°. |cosec2 60° + sec2 45°-2cot2 9. tan2 30° + 2 sin 60° +tan 45° - tan 60° + cos+ cos 60° - sin2 10. cot2 45° 11. 3 tan2 30° 60° - 30 . 1 cot^ 60°. 14. + 1 cos2 30° - h sec^ 45° - J sin'30° + cos^ 30° - sin 30 . cos 60° - tan2 45° + 1 tan^ 60° tan^ 30° + ^ sin^ 45° tan2 60°. i sin2 60° - 1 sec 60° = x sin 45° cos 45° tan 60°, If tan2 45° - cos2 15. Find 12. 13. 60°. fi .v ul .r. from the eqnation 45° cot2 30° sec 60° tan .V from a The complement of an angle defect is its right angle. Two a = -êc^ 45° cos^^lW^ sin 30° cos2 45 Definition. 37. i.s i angles are said to l)e complementary when their sum right angle. angle Thus in every right-angled triangle, each acute next complement of the other. For in the figure of the 90°. of A and C is if B is the right angle, the sum 0= 90° - ... A , and A Trigonometrical Ratios of = 90° - is the article, (l Complementary Angles. Let ABC he a right-angled triangle, of which B is the right angle then the angles at A and V are com38. plementary, so that . • . sin (90° - C=90° - A. AB /I ) = sin 0= ^^ = c.< .s .1 ; rt/f and cos (90° - Similarly, .4) it = cos C=^= sin A may tan (90° -.4) cot (90° - .4 ) be proved that = cot .4, = tan .4 j ; and sec (90°- .4) c;osec (90° = cosec -A)= sec A yl, . ) http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 45/467 . 8/12/2019 . Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 28 we If 39. [OHAP. define the co-sine, co-tangent, co-secant, as the co-functions of the angle, the foregoing results may be embodied in a single statement an angle each function of of its complement. As an which it is of illustration equal corresponding cofunction to the we may this refer Art. to 35, from will be seen that sin 60° = cos 30° = '^ sin 30° = cos tan Example 60°= cot 60° = ; ^: 30° -^/3. Find a value of A when cor 1. cos Since the equation becomes 2^ ::^ sin (90° sin (90° - 24) 90° .-. = sin 3 J 2.4 -24), = sin 34 ; -24 =34; 4 = 18°. whence Thus one value of A which satisfies the equation is 4 = 18°. In a later chapter we shall be able to solve the equation more completely, and shew that there are other values of 4 which satisfy it. Prove that sec 4 sec (90° -4) Example^. Here it will be = tan 4 + tan (90°- found easier to begin with the expression on 4). tlie right side of the identity. The second side = tan 4 + cot 4 cos A _ sin 4 sin 4 cos 4 cos 4 = sec 4 sin 4 + cos A cos 4 sin 4 sin- 4 cosec 4 = sec 4 EXAMPLES. sec (90° IV. -4 ) b. Find the complements of the following angles 1. 67° 30'. 4. 4.5° -4. 2. 25° .30 . 3. 10° '.3 . 5. 4.5° + /i. 6. 30°-^. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 46/467 . . 8/12/2019 . . . . . Elementary Trigonometry by Hall and Knight rOMPLKMENTARY ANGLES. IV.] 7. 29 In a triangle C is 50° In a triangle A is the complement of 40° and ,-1 is the complement of 10° ; findi?. 8. complement of 20° find ; Find a value of A B and ; (7. in each of the following equations 10. cos 3^4= sin 7^4. 11. tan ^4= cot 3^, 12. cot ^ = tan J. 13. cot A = tan 14. set; 5.-J 2.4 , : ^4= cos 4^. sin 9. the is ' ' - ' = cosec A Prove the following identities: = sin U' 15. sin (90° -.4) cot (90° -.4) ^ 16. si ^ 17. COS ^ 18. sin^cos(90°- J) 19. cos (90° -yl) cosec (90° -.4) 20. cosec2 (90° «, 1/ n .4 tan (90° A tan - A ) sec tan (90° - .4 \/ 23. tan2 / 24. tan (90° - J ) + cot „^ sin (90° -J) -A) 1 .sin(90°-/J)=-l. A = tan J. cosec^ (90° -A). ) A _26. cosec^ ^_ cot — -A)- siu^ ^1 sec^ (90° (90 .sec 28. sin- ^) = cot sec 27. cosec (90° - + cos/l - J)= 1 + 22. . A) - .4 ) = cot .4 A cot (90° -A) sec (90° - J) = 1. (90° - A )~coiA cos (90° - J tan (90° - .4 ^^ 217' sin v/ (90° J. j; cot (90 ^ —A) — ^—r-^ 0-^ .sec^^ o . .-I = 1. cosec (90° = ^^'' o,^„o (^0 - ^4 ) - - ,4 ). „„o .sm2(90 1. ^ secJcot^^ . .4 ) = si u A — . - ) -J) cot^ jT- (90°-^) —cosec^J ^ cos tan^ r-77^0 - J ) = cosec A (90° tan (90° cosec^ (90° .4 T-: -^) — ,- = Vtan- .4 + 1 '=--y?tAL,+.,i„(90--^). vers A ^ ' cot2j;sin2(90°-.4) 29. ^^ , ,_„ J)-cos.'l. „ A^ ^ = tan(90°sin (90° - A cot (90° - A) = cos (90° - .4 r-.-r^ cot A + cos ' 30. If 31. Find the value oi x which will satisfy .^' sec ) A cosec (90° - A)~r cot http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight (90° - .4 ) -^ ), find .r. 1 47/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 48/467 8/12/2019 Elementary Trigonometry by Hall and Knight THK USK OF IV.1 Kxiimple '1. TABI.KS. Find the value of cos 29, 26'' 28'. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 49/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 30 [cHAI'. Easy Trigonometrical Equations. We 40. now shall some examples give in trigonometrical equations. Example Find a value of A which 1. A—B 4 cos By sec satisfies the A expressing the secant in terms of the cosine, A= 4 cos 4 cos- .4 cos Since cos 30° = The student tive result in (2) Example 2. ^ we , , , 9 • COS^=^ (1), A= (2). cos or A we have = 3, C0S^= .-. equation from see - \j(1) ^^30°. that be able to understand the meaning of the negaafter he has read Chap. VIII. will Solve 3 sec Since # sec- ^ 3(1 + tan^ we have ^) 3 tan^ or = 8 tan 6-2. = 1 + tan- 6, = 8 tan ^ - 2, - 8 tan + 5 ^ = 0. a quadratic equation in which tan d is the unknown quantity, and it may be solved by any of the rules for solving quadratic equations. This is (tani?- ) (3tan6»-5) Thus tan^-l = therefore eitlier (1), 3tan^-5 = or From (1), tani9 = l, From (2), tan 5 = ^=1-6666=^ tan 59° 2' ... (2). so that 5^=45°. .-. ^ = 45°, or 59° [see Ex. 3, p. 29„], 2'. equation involves more than two functions, will usually be best to express each function in terms of the 41. it = 0, sine and When an cosine. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 50/467 8/12/2019 N< Elementary Trigonometry by Hall and Knight EASV TRIGONOMETRICAL EQUATION S. IV.] Example. We + cot 3 tan Solve 3 sin d cos d cos^ sin 6 have =5 6 cosec _ 31 d. 5 sin 6 = 5 cos 0, = 5 cos 0, 3(1- cos- ^) + 2cos2 + 5cos^-3 = 0, 3 sin^ e 4- cos2 5 cos 6' ^2cos^-l)(costf + 3)=0; therefore either From (1), cos e^-, cos^= - 2cos(?-l = 0, .(1). cos0 + 3 = O, .(2). so that ^ = 60°. a result which must be rejected as impossible, because the numerical value of the cosine of an angle can never be From (2), greater than unity. 3, [Art. 16.] EXAMPLES. Find a .solution of IV. d. each of the following equations. (Tables must be used for Examples marked witk 1. 2 3. sec = cosec 6. 6 = 4 cos 6. .sin = 3 cosec 1/5. 4 sin ^ ^7. V2cos^ = cot^. ^9. .sec2(9 /Il- sec2(9 2. tan ^ = 3 cot 4. sec 6 ^6. ^. 8. , = 2tan2(9. ,10. = 21. = 3. 2cos2(9 + 4sin2<9 = 3. 4sin^ = 12sin2^-l. tan ^ = 4 - 3 cot cos2 $ - sin2 ^ = 2 - 5 cos = 2 sec 6. cot ^ + tan 23. tan ^ - cot ^ = cosec 25. 2 sin 6 tan 26. 6 tan ^ - *27. 5 tan ^ + 6 cot 15. 17. *19. 20. (9. = 4. tan ^ = 2 sin <9. cosec2^=4cot'^^. tan2^ = sec2^ cot2^ + cosec2^ asterisk.) — cosec ^ = 0. ccsec^ ^ l2. 3tan2(9-l. ait, + 7. = l. 14. 2(cos2^-siii-^) 16. 6 cos- ^ 18. 2 22. 4cosec^ + 2sin^ = 9. 24. 2 cos ^ — H- cos 6. sin- ^-3 cos ^. (9. 6. (9 (9 ^. + 1 = tan ^ + 2 5 ^/3 sec ^ (9 + = 11. + 2^/2 = 3 sec ^. sin 6. 12 cot ^ = *28. sec''^ 6 + tan- 6 — '\ tan 6. 3-2 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 51/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETKY. 32 [CHAI'. MISCELLANEOUS EXAMPLES. of (1) sin 21 ft. made 6-4 63'' (2) ; 21' 36 . Shew that 2. 3. 25^ 37 angle i-ight a Express as the decimal 1. A. A A cos A tan A + cos A A sin cot A = 1. long just reaches a window at a height of find the cosine and cosecant of the angle from the gromid by the ladder with the ground. ladder 29 : A = —, 4. If cosec 5. Shew that 6. Keduce a = 40, 8. ABC find - cot (2) ; ^4 co.s A cosec A — 1 = 0. measure -0003 of a right angle. a triangle in which cot J, sec A, sec C. is b, sec A. B is a right angle if c' = 9, ? (1) 4sin^=l; (2) 2sec^=l; Prove th at cos ^ vers ^ (sec ^ + 1) 7 tan ^ [3) Express sec a and cosec a in terms of cut 11. Eind the numerical value of 3 tan - 30' If tana^— n 2 1 - sec 60° 5 cot- 45' - <j + +4 , find sin a and = 40. = si n- ^. 10. 12. ; Which of the following statements are possible and which impossible 9. cosec ^ ^1 75 A and find tan to sexagesimal 17« 18' y^l) 7. ft. sec . a. . . sni- 6( ) . «. m sexagesimal minutes arc eipiivalcnt to n centcsiniHl minutes, prove that ?h^-54?i. 13. If http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 52/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight MTRCKLrANKOns KXAArPLKS. TV. If 14. =- sill .( that t.ui cos (90° - A) |>rov«» , -fseo .1 33 A. .1 =3, wlieii .1 is an acute angle. cot (90'' - A) ^Q^f 16. P(tf that Shew 15. is cot .T. = -2l, PR =-20, A triangle find tan and cosec (^ Shew 18. Vind a value of 6 which satisfies sec 6^ = cosec .( ) a is = cos .1 a=\ '2 cos- the eciuation liS. tan2 60 if a. , ^::^ ' Prove that 19. angle; i-ight V- (tan a - cot a) sin a cos 17. tliat P which in - tan (90 = cot^ m - 2 tan- 45 - 2 si n- - . id cosei- - , 4. ) . -t . 20. Solve the equations 3sin^ (I) 21. l'i- \ 22. Tn e that tlie = 2cos-/9: 1 + -1 1 se.->' .( 1'^ r. tan- tan ^ - - .( sec^ sec-' ^ ^ .T t.iii^ .1 ,( - 0, equation fisin-'<9 shew that one value of 23. : in a triangle A + 4=0, impossihle, and find is ^4^C tan 1] sin<9 right-angled at A + tan B = -^ =c, shew that 24. If cot 26. Solve the equation ab c c -|- ~ ^ 3sin-<9-|-. >sinfl (', tlie other value. prove that . = sec A cosec A. = 2. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 53/467 8/12/2019 Elementary Trigonometry by Hall and Knight CHAPTER Y. SOLUTION OF RIGHT-ANGLED TRIANGLES. 42. three Every angles. triangle has six parts, namely, three sides In Trigonometry and is it usual to denote the three angles by the capital letters A, B, C, and the lengths of the sides respectively opposite to these angles by the letters a, must be understood that a, numerical number quantities b, b, c. It c are the expressing of units of length contained in the three sides. 43. We construct know from Geometry that it is always possible to a triangle when any three parts are given, provided that Similarly, if the values of one at least of the j)arts is a side. suitable parts of a triangle be given, we can by Trigonometry find the remaining parts. The process by which this is eftected is called the Solution of the triangle. The general stage ; solution of triangles will be discussed at a later in this chapter we shall confine our attention to right- angled triangles. 44. From Geometry, we know that when a triangle is right- any two sides are given the third can be found. Thus in the figure t)f the next article, where ABC is a triangle rightwhence if any two of the three angled at A we have a^ = b- + c' angled, if ; , quantities a, i.s b, c are given, the third may be determined. angles are complementary, so that Again, the two acute given the other is also known. Hence really only in the solution two cases of right-angled to be considered I. when any two sides are given IT. when one and side o?je triangles if one there are : ; acute angle are given. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 54/467 8/12/2019 Elementary Trigonometry by Hall and Knight SOLUTION OP RIGHT-ANGLED TRIANGLES. 45. Case I. To solve 3;-) a right-angled triangle when two sides are given. Let ABC be a right-angled triangle, of which angle, and suppose that any two sides are given C may then the third side A is the right A A be found from the equation rt2= 62 + ^2. Also COS 6'= - a whence C and B may Given Example. Here c^ .-. Also lso , and i? = 90° - C; be obtained. B = 90°, a = 20, h = ^0, solve the triangle. = h--a'^ = 1000-400 = 1200; = 20^/3. sin4-^-'^-i.4 -^-^^-2, sin .-. ^=30°. And C=90°-^ = 90°-30° = 60°. in The solution of a trigonometrical problem may often be obtained more than one way. Here the triangle cnn be solved without using the geometrical property of a light-augled triangle. Another solution may be given as follows ^ .-. And Also a 20 1. C=60°. A = m°-C = 90° - G0° = 30^ 4 = cos .4= COB 30° = ^; .-. c = 20V3. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 55/467 8/12/2019 Elementary Trigonometry by Hall and Knight KLKMF.NTARY TRIGONOMETRY. 3G Case 46. and one .tide a solve right-anc/1e<l truivrilc vhcD ovo. acute angle are given. ABC Let To II. [en, be a right-angled triangle is whict one A suppose of angle (' the side are given whence B, right angle, and one C, J b a, r and acute then ; y = sec. B = ^0°-<\ I b may l)e b = tan C • determined. K.riimplt' 1. Given 7>=i)0°, yl=.SO°, have T i)0° We - Also r = ; ), solve the triangle. J =90°- .W^tiO . .=tan.SO'^; a~r^ta,n SO .-. Aâin, ' = - - -=sec30 ; .5 , .-.?, = .- 2 sec 10 10^3 ono r ..O-.'Jx^^^-^.-/. each case we write down a ratio which connects the side we are finding with that whoxe value given, and a kuowledge of the ratios of the given angle enables us to complete the sohition. The student should observe that NoTK. in ?.s- Kxamjile 6' If 2. = 90°, B = 2^°4H', and r.^100, solve the triangle by means of Mathematical Tables. Here A=90°-Ii = 90°-2.5°4S' = ()4'17'. Now that ~ is, = cosB ; :----=cos 25° 43'; a = 100 cos 25° 43' = 100 X •9012 = 90-12, - Also .-. /> = 90-12 = tan from tlie Tables. U, or ^ = a tan X tan 25° 43' = 90-12 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight /> ; x •4817=^43-41. 56/467 8/12/2019 Elementary Trigonometry by Hall and Knight SOLUTION OP RIGHT-ANGI.KP TRIAXOLES. v.] EXAMPLES. (Ttihh'x mitAt 1)1' V. UT a. v.vd for Examples marked in'th an asterisk.) SdIvo tho triangles in which the following parts are given 1. t- ^3. .1=90', a C=m\ 5. r/ 7. /> = 20, ^ Z> h=V2, = < = 2, *13. /? = 90°, 15. Z? = f'=45°, r'=37°, /) = r=90°, cot If 18. Ifr'=90 , .4 = -07, .4 = 38^19', If a = 100, /> 6 = 49, /> = ;), B = m\ W=:27, C=63°, J=.30°, 7? = 60°, 2/?=f'=(50 , =90 . C'=90 . , 6 rf a 6 = 4., = 6. = 20,,'3. = 8. find a. = 50, find l9' = -(i2. c r=78 12', .4=90 , = 30, ^ . =90 , a- /i=90 , C'=40°r)l', find = 20(), = 90 /i = 6x/3, B = 90\ J 16. sin38° given *20. 2^=/> 14. r:^4. 17. Tf 8. *12. 10i>. /> «=r)^/3, *10. rA *11. *19. ^=.60, 6. ,1=90 . = 9x/3, X = 30\ vl=54°, <? = 8, 6'=90 . 9. r=6, 6 = 12, 2. t^4. = 20, ^=90. r- ('=90°, = 2^/3. a = AÎ?,. = A, : r. lind a to the nearest integer. 21. gi Tf /?= 90 , tan 36 ven 22. A =36 I f .4 = 90 , c = = 37, sin 21° 43' given *23. If *24. If *25. If , <•= 100, solve the triangle 73, a= = sec 36 1 ; = 1 -24. 00, solve the triangle 37, cos 21° 43' = -93. = 90°, B = 39 24', h = 25, solve the triangle. r/= 90°, a = 22-), h = 272, solve the triangle. C=90°, 6 = 22-75, c=2o, solve the triangle. . I be found that all the varieties of the solution of right-angled triangles which can arise are either included in the two cases of Arts. 45 and 46, or in some modification of them. Sometimes the solution of a ])roblem may be obtained by The two examples we give as solving two right-angled triangles. illustrations will in various forms be frequently met with in subsequent chapters. 47. It will http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 57/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 38 [chap. Example 1. In the triangle ABC, the angles A and B are equal to 30° and 135° respectively, and the side AB is 100 feet; find the length C upon of the perpendicular from Draw CD perpendicular produced, and Then l AD AB 135° = 45° BD = CD = x. Now in angle ADC, CD produced. CD = x. CBD = 180° .-. that let to AB the right-angled tri- = tanX>.4C = tan30°; is, .r-f-lOO ^:^' .r^3 .-. a; (^3-1) = 100 100, _100(v''3 -JS-1.-. = j;-f 100; 3' + l) 3-1 = 50x Thus the distance required is -50(,/3 + l); 2-732. 136-6 feet. Example 2. In the triangle ABC, a = 9-6 cm., Find the perpendicular from A on BC, and hence = 5-4 cm., 7? = 37°. find A and C to the c nearest degree. lu the right-angled triangle ABD, ^~ = cos ABD = cos 37° BA : BD=BA cos 37° = 5-4 X -7986 (from the Tables) = 4-31 cm. Also = sin.47?7). :sin 37°; 4^ AB AD = ABs\n 37° = 5-4 x -6018 = 3-25 But CD = Tie - 7,'7) cm. = 9-6 - 4-31 from the right-angled triangle (from the Tables) =5-29 cm. ACD, http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 58/467 8/12/2019 Elementary Trigonometry by Hall and Knight SOLUTION OF RIGHT-ANGLED TRIANGLKS. v.] 39 But from the Tables, tan 31° = -6009, / .-. ^ CD = 32°, tan 32° = -6249 to the nearest degree, /B^C=180=-37°-82°=lir. iA = lir, / C = 32°. and Thus ^D=:3-25cm., EXAMPLES. V. 1. dnced : ABC find a triangle, and BD, given is .4 =-30°, 2. If BD a and c, is asterisk.) perpendicular to AC pro- AC =20. C=120°, AC oi a, triangle ABC, given =30°, 0=45°, In the triangle making BD equal to 15 3. B is perpendicular to the base J given that BD b. marked with an (Tahles mnst he used for Examples find ; and C ABC, ft. : AD find are equal to BD = \0. drawn perpendicular to BC the lengths of A B, AC, and A D, 30° and 60° respectively. is In a right-angled triangle PQR, find the segments of the hypotenuse PR made by the perpendicular from Q given 4. ; LQRP=m\ QR = S, *5. If PQ given find RS, *6. find If PQ RS, given drawn perpendicular is = Zb% i drawn perpendicular PQ = 20, to the straight line i PRS =135% QRS, SPQ= J'i°. l RPQ PQ=ZQ, is lQPR=-ZQ\ to the straight line l QRS, PSR^'lb . the angles B and C are equal to 45° and 120° respectively; if a = 40, find the length of the perpendicular from A on BC produced. 7. *8. find If CD DC and is drawn perpendicular BD, In a triangle lîerpendicular to the straight line DBA, given .45 = 41-24, *9. A BC, In the triangle from A zCi?i)=45°, ABC, i C^ 27 = 35° AB = 20, BC =33, on BC, and the angle http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight i 7? 18'. = 42°, find tlie C. 59/467 8/12/2019 Elementary Trigonometry by Hall and Knight CHAPTER VT. EASY PROBLEMS. 48, now principles explained in the lirevioiis chapters to the solution of problems ifi heights The be applied may and be assumed that by the use of suitable instruments the necessary lines and angles can be measured with sufficient accuracy for the purposes required. distances. After It the will practice afforded by the examples in the last chapter, the student should be able to write down at once any side of a right-angled triangle in terms of another through the of the functions of either acute angle. In the present impoi'tance to acquire it is of great medium and subsequent chapters readiness in this respect. Fur instance, ftvjm the adjoining figure, A we liave C <5 a = c sin A, a = c cos B, a = h cot B, a = biax\ A, ct=aseci?, 6 = atan^. committed to memory but in each case should be read off from the figure. There are several other similar relations comiecting the parts of the above triangle, and the student should practise himself in obtaining them These relations are not to be quickly. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 60/467 8/12/2019 Elementary Trigonometry by Hall and Knight 41 KASY TROBLEMS. a'ld '^'^' i« three points in a straight line, V, H, T are iPQT-^, If PTâ, drawr^ perpendicular to QT. terms of a and /I lines of the figure express the lengths of all the Ktamnlf. m '/,VnKT I. ByEuc, 32, .-. ^^^ lQPR=^lPliT- lPQB; iqPR^2^-^=^-- iPQi^; .-. QR=-PR. In the right-angled triangle PR'J\ PR = a cosec .-. QR = TR = a Also (T i/S cosec 'ip. cot 2^. PQT, Lastly, in the right-angled triangle QT — a cot p, PQ = a cosec/?. Let OP Angles of elevation and depression. plane as an oh.iect (^, and horizontal line in the same vertical be a 49. 00 ^ be 'ioined. let P O Fig.1 ; angle of -J^pression -\,o Fig.2 angle of elevation aborc the horizontal line Ol\ in Fig. 1, where the Q of elevation of the object the angle P0§ is called the angle object is as seen from the point 0. horizontal line Ol\ where the object Q is beUnc the of depression of the object q the angle P0(? is ca,lled the angle as seen from the point 0. In Fig. 2, http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 61/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 42 Example plane, on a horizontal and from a point on the ground at a distance of 30 its A I. flagstaff stands angle of elevation ft. its height. AB be the flagstaff, observation; then Let of 60°: find is AB = BC tan = 30 Thus the height C 60°:= 30 the point sJ-A X 1-732 ôl-96. is 51 '96 ft. EXAMPLES. VI. [/by Examples in the une of 1. The angle From depression a 7'ublex see of elevation of the top is 30° find its height. distance of 300 feet 2. a. page 48x.] of a chinuiey at a : ship's of a boat masthead is 160 observed to feet high be 30° : the find its angle of distance Irom the ship. 3. Find the angle of elevation of the sun when the shadow of a pole 6 feet high is 2 v'S feet long. At a distance 86 '6 feet from the foot of a tower the angle of elevation of the top is 30°. Find the height of the tower and 4. the observer's distance from the top. A ladder 45 feet long just reaches the toji of a wall. If the ladder makes an angle of 60° with the wall, find the height of the wall, and the distance of the foot of the ladder from the 5. wall. 6. Two masts joining their tops are 60 feet and 40 feet high, and the line makes an angle of 33° 41' with the horizon find their distance apart, given cot 33° 41' 7. : =1 5. Find the distance of the observer from the top of a cliff is 132 yards high, given that the angle of elevation which 41° 18', and that sin 41° 18' = -66. is is 30 yards higher than another. A person standing at a distance of 100 yards from the lower observes their tops to be in a line inclined at an angle of 27° 2' to the horizon: find their heights, given tan 27° 2' = 'SI. 8. One chimney http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 62/467 . 8/12/2019 : Elementary Trigonometry by Hall and Knight EASY PROBLEMS. VI.1 43 From the foot of a tower the angle of elevation of the top of a column is 60°, and from the top of the tower, which is find the height of the 50 ft. high, the angle of elevation is 30° example II. : column. Let to DB. AB denote the column and CD the tower ; draw CE parallel IjetAB = .v; AEÂB-BE = .v-oO. then DB=C'Eî/. Let From the right-angled triangle 7/ From = .z;cot60°=-^. the right-angled triangle y = {x - 50) cot 30° .-. = ^3 {x ACE, - 50) ~3 = ^3(.c-50), a; whence = 3(.r-50); ;c=75. Thus the column The angle 9. ADB, is 75 ft. high. of elevation of the top of a tower walking 100 yards nearer the elevation the height of the tower. A is is 30 found to be 60° ; on find stands upon the top of a building.; at a distance of 40 feet the angles of elevation of the tops of the flagstaff and building are 60° and 30° find the length of the 10, flagstaff : 11. of it of elevation of a spire at two places due east feet apart are 45° and 30° find the height of the The angles and 200 : spire. 12. From steeple is 45°, the foot of a post the elevation of the top of a and from the top of the post, which is 30 feet high, the elevation is 30°; find the height and distance of the steeple. 13. The height of a hill is 3300 feet above the level of a horizontal plane. From a point A on this plane the angular elevation of the toja of the hill is 60°. balloon rises from A A and ascends vertically upwards at a uniform rate ; after 5 minutes the angular elevation of the top of the hill to an observer, in the balloon is 30° find the rate of the balloon's ascent in miles per houi'. : http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 63/467 , 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 44 [CHAV. Example III. From the top of a cliff 150 ft. high the angles of depression of two boats which are due South of the observer are 15^ and 75^ find their distance apart, having given : cotl5'' OA Let = 2 + ^3 and represent the horizontal line through cliff, ; Let From and then = 150 cot 75° Thus the x- i OB A = 75°. = 150 OB A (2 - J3) = 300 - 150 ^'3. = 300 -f 150 ^'3. = 300 ^'3^ 519-6. distance between the boats From be a /POB = 75°; the right-angled triangle OCA, x + y = 150 cot 15° = 150 (2 + J3} subtraction, OP Let CA = x + y. the right-angled triangle y and C the boats. OCA = 15° and CB = x, AB = y; From By jL -^3. then /POC= 15° .-. B cot TS'^-^'i is 519*6 ft. mouument 100 feet high, the angles of depression of tvro objects on the ground due west of the monument are 45° and 30° find the distance between them. 14. the top of a : 15. angles of depression of the top and foot of a tower The monument 96 feet high are 30° and 60° find the seen from a : height of the tower. From 150 feet high the angles <if depression of two boats at sea, each due north of the observer, how far are the boats apart are 30° and 15° 16. the top of a cliff '{ : From the top of a hill the angles of depressiofi of twr» consecutive milestones on a level road running due south from 17. the observer are 45° and 22° respectively. the height of the hill in yards. If cot 22° ^2-475 find From the top of a lighthouse 80 yards above tlie horizon the angles of depression of two rocks due west of the observer — '268 are 75° and 15°: find their distance apart, given cot 75° 18. and cot 15° = 3-732. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 64/467 8/12/2019 Elementary Trigonometry by Hall and Knight THE MARINERS COMPASS. VI.] 45 Trigonometrical Problems sometimes require a knowledge of the Points of the Mariner's Compass, which we shall 50. now explain. In the above figure, it will be seen that 32 points are taken at equal distances on the circumference of a circle, so that the arc between any two consecutive points subtends at the centre of the circle an angle equal to ^^-°, that is to llJ^ points North, South, East, West are called the Cardinal Points, and with reference to them the other points receive their names. The student will have no difficulty in learning these The if he will carefully notice the arrangement in any one of the principal quadrants. 51. Sometimes a slightly different notation is used; thus N. 11|° E. means a direction 111° east of north, and is therefore Again ixW. by S. is .3 points from south the same as X. by E. and may be expressed by S. 33|° W., or since it is 5 points from In each of these we.st it can also be expressed by W. 56|° S. cases it will be seen that the an^ûlar measurement is made from the direction which is first mentioned. H. K. E. T. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 65/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 40 The angle between the 52. directions of [chat. any two jôints is obtained by multiplying 11^° by the number of intervals between the points. Thus between S. by W. and W.S.W. there are 5 and the angle intervals there ai-e between N.E. 56J° 7 intervals and the angle is 78|°. B If 53, lies is ; by E. and S.E. in a certain direction with respect to A, A it is thus Birmingham becu's N.W. of London, and from Birmingham the hearing of London said to hear in that direction from is ; S.E. Example. a lighthouse L two S.W. and 15° East of South in directions B From 1. .ships A and B are observer respectively. At the same If JjA is 4 mile.s find observed from A in a S.E. direction. the distance between the ships. time is Draw due South; then the bearings of the two ships, Ji,S A LS' = 45°, Z so that I BLS' = 15°, l ALB = 60°. a line Through A drawthen North and South; lNAL= and /.BAS = 45°, from A NS /.ALS' = i5 B since from pointing , bears S.E. ; hence / ii.J L = 180° - 45° 45 = 90 . In the right-angled triangle ABl,, AB = AL tan A LB ==4 tan 60° -=4^3 = 6-928. Thus the distance between the is <l-928 ships miles. Example 2. At a.m. a ship which is sailing in a direction 37° E. S. at the rate of 8 miles an hour ob.serves a fort in a direction 53° North of East. At 11 a.m. the fort is observed to bear N. 20° W.: ',) find the distance of the fort Let A and i' be the first from the ship at each observation. and second positions of the ship; B tlie fort. Through A draw From lines the observations / KACr=31 , towards the cardinal points of the compass. made / EAB.-^5-6', so that http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight / BAO = W. 66/467 8/12/2019 Elementary Trigonometry by Hall and Knight THK MARrNKR's COMPASS. VT.] Through C draw CN' towards the North; then the bearing of the fort from Also A .-. f ' i« iACB= lACN' C tan A (W = 1 fi -^ l T{CN'-20'', for W. A CN' = iCA S = 90 In the right-angled triangle AR-^.4 N. 20 - 37'' = 53° /BCiV' = 53° -'20 = 33 . ACE, tan 33° = 1(^ < -6494, from the Tables ; = 10-3904. Again nCÂC see ACB = IC. see 33° - 16 x 1-1924 Thus the distances are 10-39 and 1^-08 miles EXAMPLES. {For Exampfr/t to hf salved = 190784. nearly. b. VI. by the use of Tables see page 48^.) A person walking due E. observes two objects both in the After walking 800 yards one of the objects is due N. of him, and the other lies N.W. how far was he from the objects at first ^ 1. N.E. direction. : Bailing due E, I obsei've two ships lying at anchor due S. 2. how after sailing 3 miles the ships bear 60 and 30 S. of W. ; far are they 3. Two now distant from me ? noon in directions W. 28° S. at the rates 10 and lOi miles per hour respectively. vessels leave harbour at and E. (52 ' S. Find their distance apart at 2 P.M. 4—2 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 67/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 48 A 4. lighthouse facing N. [CHAP. VT. sends out a fan-shaped beam steamer saihng due W. first A extending from N.E. to N.W. sees the hght when 5 miles away from the lighthouse and What is the speed of the continues to see it for 30,^/2 minutes. steamer 1 A 5. ship sailing due S. observes two lighthouses in a line W. and W.N.W. exactly After sailing 10 miles they are respectively N.W. find their distances from the position of the ship ; at the first observation. Two vessels sail from port in directions N. 35° W. and W. at the rates of 8 and 8^/3 miles per hour respectively. 6. S. 55° Find their distance apart at the end of an hour, and the bearing of the second vessel as observed from the first. A noon to be E.S.E. At 1 p.m. the vessel is due S. from a lighthouse 4 miles away. find the rate at which the vessel is sailing. of the lighthouse Given tan67|° = 2-414. 7. vessel sailing S.S.W. is observed at : A, B, C are three places such that from A tlie bearing of from B the is N. 50° E. is N. 10° W., and the bearing of C between B and C is If the distance bearing of C is N. 40° W. B 10 miles, find the distances of B and C from A. 8. ; A ship steaming dixe E. sights at noon a lighthouse bearing N.E., 15 miles distant; at 1.30 p.m. the lighthouse bears N.W. How many knots per day is the ship making ? Given 9. 60 knots = 69 miles. 10. At 10 o'clock forenoon a coaster is observed from a lighthouse to bear 9 miles away to N.E. and to be holding a .south-easterly course; at 1 p.m. the bearing of the coaster is 5° Find the rate of the coaster's sailing and its distance from the lighthouse at the time of the second observation. 1 S. of E. vessel is which N.E. of A is sailing S. 15° and N.W. of B A and />, is 12 At midnight a E. at the rate of 10 miles per hour find to the nearest minute when The distance between two lighthouses, miles and the line joining them bears E. 15° N. 11. : the vessel crosses the line joining the lighthouses. From A to B, two stations of a railway, the line runs W.S.W. At ^ a person observes that two spires, whose distance apart is 1-5 miles, are in the same line which bears N.N.W. At B their bearings are N. 7^° E. and N. 37^° E. Find the rate of a train which runs from A to B in 2 mimites. 12. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 68/467 8/12/2019 Elementary Trigonometry by Hall and Knight EASY PROBLEMS. VI.] EXAMPLES. 48^^ VI. c. [In the foUoioing Examples Four- Figure Tables vdll be required.'] At a 1. distance of top of a chimney stack is 8.3 yards the angle of elevation of the find its height to the nearest 23° 44' ; foot. The shadow cast by a 2. elevation of the sun in 63° spire at a time is when the angular find the height of the 173 feet; spire. balloon held captive by a rope 200 metres long has wind till the angle of elevation as observed from How high is the balloon above the the place of ascent is 54 . * 3. A drifted with the ground -• ? A 4. zontal. a man where coal seam is inclined at an angle of 23 to the hori- Find, to the nearest foot, the distance below the level of who has walked 500 yards down the seam from the point it meets the surface. A man standing immediately opposite to a telegraph post on a railway notices that the line joining this post and the next Assuming that there are 23 one subtends an angle of 73° 18'. telegraph posts to the mile, find his distance from the first post. 5. The middle point 6. of one side of a square of the opposite corners of the square; is joined to one find the size of the two angles formed at this corner. Find without any measurement the angles of an isosceles triangle each of whose equal sides is three times the base. 7. * of it, of elevation of a spire at two places due east 160 feet apart, are 45° and 21° 48': find the height of The angles 8. and the spire to the nearest foot. ' 9. The angle of elevation of the top of a tower is find the height of the foot. 10. From the top of a hill 27° 12', found to be 54° on walking 100 yards nearer the elevation tower to the nearest is and 24' the angles of depression of two consecutive milestones on a level road running due east from the observer are 55° and 16° 42' respectively; find the height of the hill to the nearest yard. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 69/467 8/12/2019 Elementary Trigonometry by Hall and Knight Kl.EMENTAKV TRKiONOMKTKY. 48 [OHAT. VI. A Irouch is to be dug in measure 1 ) feet across the top, if one side I) feet across the bottom, and of uniform depth 8 feet 12° inchnatioii of to the vertical, what must be the is incHned at 11. ; tiie other '( Two and B, on a level plain subtend an augU; at an observer's eye; he walks directly towards B a of 90 distance of 630 metres and then finds the angle subtended to Find the distance of A from each position of the be 143° 24'. 12. towers, î observer. From the roof of a house 30 feet high the angle of elevation of the top of a monument is 42°, and the angle of 13. depression of its foot is 17°. 14. From 15. At noon a ship which Find its height. a point on a horizontal plane I find tlie angle of elevation of the toj) of a neighbouring hill to be 14°; after walking 700 metres in a straight line towards the hill, I find the Find the height of the hill. elevation to be 35°. is due W. sailing a straight course At at 10 miles an hour observes a lighthouse 32° W. of North. find the distance 1.30 P.M. the lighthouse bears 58° E. of North of the lighthouse from the first position of the ship. ; 16. From two road running East and 38° E. of N. respectively. house is from the road. 17. A 18. A 2 kilometres apart, on a str-aight West a house bears 52° W. of N. and positions, Find to the nearest metre how far the ship sailing due E. observes that a lighthouse known to be 12 miles distant bears N. 34° E. at 3 p.m., and N. 56° W. at 4.10 P.M. How many miles a day is the ship making ? ship which was lying 2h miles N.W. of a shore range of 4 miles, steers a straight course inider cover of darkness initil she is due N. of the battery, In what direction does she steam ? and just out of range. battery with an effective which is sailing E. 41° S. at the rate At of 10 miles an hour observes a fort bearing 49° N. of E. 19. At 10 A.M. a ship noon the bearing of the fort is N. 15° E. the fort from the ship at each observation. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight ; find the distance of 70/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 71/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. [CHAI . )2Vie circuraferences of circles are to one another as their 57. Take ajiy circle let two circles whose radii are r^ and rj, and in each a regular polygon of n sides be described. Let xi^Bi be a side of the first, J 2 ^2 ^ ^^^^^ ^^ polygon, and let their lengths be denoted by aj, a.^. second Join their We thus extremities to 0^ and 0^ the centres of the circles. obtain two isosceles triangles whose vertical angles are equal, each ^''^ being - of four right angles. Hence the by Euc. triangles are equiangular, VI. 4, cp7 ~ tliat and therefore we have 0^'^ ' ii- '2 na., that is, r, r. where p^ and jOj are the perimeters of the polygons. This is true By taking whatever be the number of sides in the polygons. n sufficiently large we can make the perimeters of the two differ from the circumferences of the corresponding by as small a quantity as we please so that ultimately polygons circles whei'e ; t'l and c.^ are the circunifei'ences of the two circles. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 72/467 8/12/2019 Elementary Trigonometry by Hall and Knight RADIAN MEASURE. VII.] It thus appears that the ratio of the circumference of 58. / ^circle that 51 to its radius is the same whatever a he the size of the circle; is, „ . circumference , m all circles . is j^-^ diameter .-. . a constant quantity. always denoted by Though its numerical value cannot be found the Greek letter n. exactly, it is shewn in a later part of the subject that it can be obtained to any degree of approximation. To ten decimal places 22 In many cases 17=-^, which is true its value is 31415926536. This constant is incommensurable and is two decimal places, is a sufficiently close approximation where greater accm-acy is required the value 3-1416 maybe used. to 59, If c denote the circumference of the circle whose radius we have is r, circumference diameter c=2nr. To prove that Draw any circle ; all radians are equal. be let its centre be radius. Let the arc and OA a equal Join in length to OA. AB measiu-ed OB A OB a radian. Produce ^0 to meet the circumference in V. By Euc. VI. 33, angles at the centre of a ; then z circle is are proportional to the arcs on which they stand ; hence lAOB arc two right angles ~ which to is ABC radius _ semi-circumference ' irr _ ^ n ' a radian always bears the same and therefore is a constant angle. constant; that two right angles, arc AB is, http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight ratio 73/467 8/12/2019 Elementary Trigonometry by Hall and Knight KLEMENTARY TRIGONOMETRY. Since a radian [CHAI*. taken as a standard unit, and the number of roAians contained in any angle is spoken of as its radian measure or circular measure. [See Art. 71.] In this system, an angle is usually denoted by a mere number, 61, is constant it is the unit being implied. Thus when we speak of an angle 2*5, it is understood that its radian measure is 2 '5, or, in other words, that the angle contains 2^ radians. ^ To find 62. the radutii measure of a riijhi. itngle. A OC Let of a circle, be a right angle at the centre and aAOB a radian then ; the radian measure of lAOG I AOB -- i A OC arc/lC AB arc - (circumference) (27rr_) [ad ins that is, a rij^ht angle contains - radians. To find 63. Fiom the number of degrees the last article 7T- radians it =2 a radian. follows that right angles =^180 degrees. .. .•. in, a radian— 180 , degrees. IT By division we find that - = 'SlBSl hence approximately, a radian 64. The formula IT = nearly 180 x -31831 ; = 57 2958 degrees. radians = 180 degrees connecting the sexagesimal and radian measures of an angle, is a useful result which enables us to pass readily from one system to the other. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 74/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight 53 HAUIAN MEASURK. VIl.] Express 75 Example. in radian measure, and . . iu sexagesimal measure. Since 180 degrees — r radians, (1) Thus the radian measure =180 J. ~- radians oi = ^- TT TT - is radians Since (2) radians _r— radians. ^TT 75 degrees— 180 . degrees, ^ degi'ees. 54 — degrees = 3° Thus the angle = 20'. be well to reniiud the student that the .symbol n When the symbol always denotes a number, viz. 3'14159.... stands alone, without reference to any angle, there can be no ambiguity but even when n is used to denote an angle, it must still be remembered that tt is a numher, namely, the number of may It 65. ; radians in two right angles. Note. such as mode It is not 7r = 180 uncommon or ^= for beginners to make statements Without some modification 90. of expression is quite incorrect. It is true that ir this radians are is no more correct equal to 180 degrees, but the statement '7r = 20= 1 to denote the equivalence of 20 shillings than the statement 180' and 1 sovereign. ^%^^J^-tke number of degrees and roAians in resented hy and 6 respectively, to prove that an antjlc he D 180 tt In sexagesimal measure, the ratio of the given angle to two right angle., , expressed by ,4. In radian measure, the ratio of these same two angles is expressed by IT • 180 ~7r http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 75/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 54 Example 1. What [CHAP. the radian measure of 45° 13' 30 is be the number of degrees in the angle, have D = 45 -225. ? If I) we Let 6 be the angle, then e _ _ ~ 45-225 7r~ 180 ^=.^ X 1-005= 4 the radian measure Example 2. radian measure Let D be the 30 60 ) 13-5 1-005 4 is ' --^^x 4 = •7854x1-005= Thus ) of radians in the given number .-. 60 1-005 -789327. -789827. Express in sexagesimal measure the angle whose is 1-309. number of degrees D ; then 1-809 180 TT ' 180x1-309 180x1309x10 3-1416 31416 180 X 10 24 Thus the angle is 75°. EXAMPLES. VII. [Unless otherwise stated It 7r a. = 3-1416. should be noticed that 31416 = 8 x 3 x 7 x 11 x 17. Express in radian measure as fractions of n 1. 45°. 2. 30°. 5. 18°. 6. 57° 30'. : 3. 105°. 4. 22°^ 30'. 7. 14° 24'. 8. 78° 45'. Find numerically the radian measure of the following angles: 9. 25° 50'. 10. 37° 30'. 11. 82° 30'. 12. 68° 45'. 13. 157° 14. 52° 30'. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 30'. 76/467 8/12/2019 Elementary Trigonometry by Hall and Knight RAPTAN MKASURR. VII.] Express in sexagesimal measure % 15. ^. 4 16. 19. -3927. 20. Taking 27. 22 7r = ^, : 17. 45 6720. 23. 36° 32' 24 . 25. 116° g. 27 21. find the radian 55 18. ^. 24 22. 2-8798. -5201. measure of: 2' 45-6 . 24. 70° 33' 36 . 26. 171° 41' 50-4 . =-31831, shew that a ividian contains 206205 Taking TT seconds approximately. 28. Shew that a second is approximately equal to -0CHX)048 of a radian. 67. Tlie angles measure of the angles Hence the f( 7, o' ;^ ^^^ ^^^ equivalents in radian 45°, 60°, 30° respectively. results of Arts. 34 and 35 may l>e written as )llows . TT 1 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 77/467 8/12/2019 Elementary Trigonometry by Hall and Knight KI-KMENTAKY TRIUONOMRTRY. 5fi When 68. of 6 is — g now have expressed radian in and corresponding ^, measiire [CHAP. complement the to the formulae of Art. 38 we relations of the form sin ( ^- <9 ) 2 = <•<)« tan d, ^ ( - ^ = cot ) rf. Prove that E.nimple. (cot 6 The tan I first sirle cot d) = (cot ^ ^ ~ | — cosec- j + tan ^)tan d — cot 6 tan + tau- 6 = 1 + tan- d = secfl = cosec'^ By means 69. of Euc. i. J-)number 32, it is easy to find the of radians in each angle of a regular polygon. Example. Express in radians polygon which has n sides. The sum of the exterior angles be the Tjpt H number nd But interior angle — mi of radians in = '2TT, and — two IT au exterior angle ~ - a 1 1. 3. tlie - . 3'2 Cor. | then . M IT n (>1-'2)ti-^^ VII. b. numerical value of Sin- cos - cot 3 b 4 i ; i. ref<nlar 'lir therefore EXAMPLES. Kind a |Enc. right angles. =— of right angles - exterior angle I angle = Tims each mtenor I =4 angle interior the cos a 3 . + 2 cosec „ . 2, tiui 4. '2 () http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight -. n sm cot ~ cos -, 3 - 4 + 4 - sec - 2 , 4 78/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight RADIAN MK.AHURK. Yjx.l Find the numerical value of cot2|+4cos2j+3sec2?. 5. 6. 7. -A tan2 ttI - ( sin ^ + cos ^ „7rl - - .sin2 I - V' Vy ) ^ sin - ( ^ ''*^*' - cos Prove the following identities ^ (| / \ ^ ^ sec2^ + se.c^ flS; ^ ) - seit (1) 15. = (^ - S\ = (1 = 0. -^ (^^ ^tan^^) sec^ (| -^ exterior angle of Find the number of radians in each a re.gular quindecagon. ('2) Kind the number of radians in a regular dodecagon, each interior angle of (±) a regular hei.tago.i. W -- COS Shew that i, . oTT tail- - - ôT cot- ;; = Shew .. 3 _ - • COS'' - that the simi of the squares of sin equal to > TT 6 COS'' 16. 1 - COS- is .^ - ^) ^ sec n) a regular octagon, 1^. ' : - 12. ^^ 6 (l ^ tan 6 + tan (^ ' sec j '- ^ ^) 11. ''^' '•<'+'^ 4 - „rr ''''' 3 sin^^J-^^cosec^-tan'^f^-^jsin^ 9. Mn 1/ ^ - sin 6 sec 8. <,t4 + (9 + sin (| - ^) ^^^^l ^'o** ^ - '^'^^ (| ^) -1. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 79/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 80/467 ; 8/12/2019 Elementary Trigonometry by Hall and Knight KAUIAN MEASURE. Vil.] 59 In running a race at a uniform speed on a circular course, a man in each minute traverses an arc of a circle which subtends 2f radians at the centre of the course. If each lap is 792 yards. Example how long Let )• 2. does he take to run a mile r tt ? yards bo the radius of the circle 22 =^ ~- . then ; = circumference =: 792 27rr 792 X 7 792 ^ 12ti. Let a yards be the length of the arc traversed in each minute then from the fornuila a = rO, (( that is, tlic i.>A = 12b man runs 360 2i X or. '. Thus the time is Example 3. measured along whose the time = 1760 -777obO 44 Find the radius of a globe such that the distance surface between two places on the same meridian its by 1° 10' may be 1 Let the adjoining figure represent a which the two places Now but F and Q inch, reckoning that 7r= T^ P(^=l .: - inch, and iP()Q = = number r of radians in z 22 1 <> /• POQ; r 10'; I 11 180 540 -^t11 = 4V,. Thus the radius H. K. . of radians in If '^^''180-6 whence - on Let lie. 22 sec- and denote the radius by arcPO — = number — radius arc ^ 9 tion of the globe through the meridian inches. . —minutes. or 4 min. 534 sec. latitudes differ bo the centre, ; yds. in each minute. ^, .'. X 20 — 126 —— = 3bO 1;. is 49,'^ inches. T. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 81/467 8/12/2019 Elementary Trigonometry by Hall and Knight 60 JELEMENTARY TRIGONOMETRY. EXAMPLES. arc of An angle whose circular measiu'e centre of a circle an arc of 219 feet 3. What 6 7 5 feet ; feet. tlie find the radius of the circle. what is is 2 5 yards the angle? the length of the arc which subtends an angle radians at the centre of a circle whose radius is 1*625 of ; 24 subtends at 73 is is angle at the centre of a circle whose radius subtended by an arc of 4. 3 An whose radius at the centre of a circle 1 '6 _;?ards 2. is c. Find the radian measure of the angle subtended by an 1. * VII. [cHAP. is ? yards An. arc of 17 yds. 1 ft. 3 in. subtends at the centre of a circle an angle of 1 '9 radians find the radius of the circle in 5. ; inches. 6. how The flywheel of an engine makes 35 revolutions in a second; 22 long will 7. The inches does 8. A it take to turn through 5 radians large its hand of a clock extremity move in 20 horse is tethei'ed 2 is ft. 4 in. mmutes ? how many long; 22~ r ? how long must has moved through 52 to a stake the rope ; 36 }'ards be in order that, when the horse at the extremity of the rope, the angle traced out by the rope may be 75 degrees I 9. Find the length of an arc which subtends 1 minute at the centre of the earth, supposed to be a sphere of diameter 7920 miles. 10. Find the number of seconds the centre of a circle of radius 11. Two places on the 1 the angle subtended at mile by an arc 5^ inches long. in same meridian apart; find their difterence in latitude, taking 145-2 are 22 7r = ^. , miles and the earth's diameter as 7920 miles. Find the radius of a globe such that the distance measured along its surface between two places on the same meridian whose 22 latitudes differ by i.^^° may be 1 foot, taking 7r= „ 12. . http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 82/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight MISCKLLANEOUS EXAMPLES. VU.] 61 B. MISCELLANEOUS EXAMPLES. Express 1. B. degrees the angle whose circular iiicasure iii is •1.5708. A =30% c^[ 2. If C'=90 , 3. Find the number of degrees two decimal lU, tiiid b to in the unit angle places. when the 1 2«j. angle is represented by What 1^. the radius of the circle in which an arc of subtends an angle of 1' at the centre 4. is 1 inch '? Prove that 5. + cos a) (tan a + cot a) = sec o + cosec - cot 30°) = tan3 60° -2 sin 60 . (v'3 + 1)(3 (sin a (1) 7 (2) 6. 60 feet (1) ,_, (2) (tan 1+- -, (9 + 2) (2 1+ cosec a express the third of +1 ) --= .-> tan $ +2 sec- 6 ; = cosec a. triangle is 45 and another is - railians; 8 angle both in sexagesimal and radian measure. The number 9. a, : tan ^ cot^ a One angle 8. ; Find the angle of elevation of the sun when a chinniey high throws a shadow 20 ^^3 yards long. Prove the identities 7. a of degrees in an angle exceeds 14 times the iiunaber of radians in it by Taking n .51. 22 =~ ^ find the sexagesimal measure of the angle. 10. irom C 5 = 30°, 6'— 90°, 6 on the hypotenuse. 11. If Shew = 6, find a, and the 2>erpendicular c, that (1) cot ^ + cot (- - ^ (2) cosec- 6 -f cosec- ( j — cosec ^ ' ^ ) ^ cosec — ^<Jsec-' I ^^^); d cosec- (--- http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight \ 83/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 84/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight MISCKLLANEOUS EXAMPLES. VII.] Prove the identities 22. +r sin a (1) ' cota (2) cosec a (sec a ^ .. 1+cosa . Shew that 23. — =COSeca: 1 ) — cot a (1 — cos a) — tan a — sin a. 1+ cos 30° /l+cot60 \2 ^ - cot 6O7 A man walking X. W. 24. : ' 1 63 B. 1 - cos 30° sees a windmill which bears X. 1 . )° \V. In half-an-hour he reaches a place which he knows to be W. Find his rate of walkof the windmill and a mile away from it. 15° S. intC and his distance from the windmill at the first observation. 25. Find the number of radians in the complement of 26. Solve the equations - 8 : 3sin^ + 4cos2(9 = 4i; (1) 27. — If 5 tan a = 4, tan (2) 6J + sec 30° = cot (9. find the value of 5 sin a — 3 cos a sin a 4-2 cos a 28. Prove that — sin A cos A cos A (sec A — cosec A) 1 29. which is 77 30. sin- J^ — cos-^ ^\v?A + cosM sin J. Find the distance of an observer from the top of a is 26', 195'2 yards and A high, given that 26' 77° '976. that sin horse cliff the an£,de of elevation = is tethered to a stake by a rope 27 feet long. If the horse moves along the circumference of a circle alway.s keeping the rope tight, find how rope has traced out an angle of have gone when 22 far it will 70°. r tt tlio = -^- http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 85/467 8/12/2019 Elementary Trigonometry by Hall and Knight CHAPTER VIII. TRIGONOMETRICAL RATIOS OF ANGLES OF ANY MAGNITUDE. In the present chai^ter we shall find it necessary to take account not only of the magnitude of straight lines, but also of the direction in which they are measured. 72. be a fixed point in a horizontal line XX\ then the position of any other point P in the line, whose distance from is a given length a, will not be determined unless we know on which side of the point lies. Let P o ;<' X But there will be no ambiguity if it is agreed that distances measured in one direction are positive and distances measured in the opposite direction are negative. Hence the following Convention of Signs lines measured frovi to the lines weaswedfrom to the left Q X' Thus in the above figure, A'X' at a distance a from bv the statements line OP=-â, 73. A : are negative. X P P adopted right are positive, O if is and Q are two points on the 0, their positions are indicated OQ=-a. similar convention of signs is used in tlie case of a ])lane surface. be any point in the plane draw two straiglit through lines XX' and in the horizontal and vertical direction re.spectively, tlnis dividing the ])laiie into fuui' (juadrants. Let YV http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight ; 86/467 8/12/2019 Elementary Trigonometry by Hall and Knight CONVENTION OF SIGNS. Then (1) it is universally agreed to consider that horizontal lines to the right of horizontal lines to the left of (2) 05 vertical lines above vertical lines below XX' XX' TY' FT' are positive, are negative are positive, are negative. Y Ms http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 87/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 88/467 8/12/2019 Elementary Trigonometry by Hall and Knight TRIGONOMETRICAL RATIOS OP ANY AXOLK. nu.] From (u seen that any trigonometrical function will be positive or negative according as the fraction which expresses its value has the numerator and denominator of the same sign or of opposite sign. these definitions The four diagrams 76. be will it of the last may article be con- venientlv included in one. With and fixed radius let a circle be descril)ed A' and YY' divide the circle into four tlien the diameters A' qaadrants XOY, YOX\ X'OY', Y'OX, imvaed Jirxt, second, thinf, centre ; fourth respectively. Let the positions of the radius vector in the four quadrants be denoted by OP^, OP^, OP^, OP^, and let perpendiculars then it will be PiJiTi, P.,M.^, P3J4, Pjfi be drami to A'A seen that in the first quadrant all the fines are positive and there; fore all the functions of In is A are positive. the second quadrant, negative ; hence sin A OP.^ is and aie positive, 021., M.,P., positive, cos A and tan A axe negative. In the third quadrant, negative; hence tan is J OP^ is positive, is positive, sin ^4 In the fourth quadrant, OP^ and negative; hence coh A is positive, Oil/3 and J/3P3 are and cos ^4 are negative. 0M^ sin M^P^ tan J are are positive, .1 and neiiative. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 89/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 68 [chap. The following diagrams shew the signs of the trigonometrical functions in the four quadrants. It will be sufficient 77. to consider the three principal functions only. tangent cositte + + + The diagram below exhibits the ful form. sine same results in anotlier use- positive the all cosine negative ratios positive tangent negative tangent positive negative sine cosine When 78, cosine an sine negative tangent negative negative angle positive increased is or diminished by any back multijile of four right angles, the radius vector is brought position after one or revolutions. into the same more There are thus an infinite number of angles which have the again same boundary line. Such angles are called coterminal angles. any integei-, all the angles coterminal with A may be represented by n. 360° + ^4. Similarly, in radian measure all the angles coterminal with Q may be represented by inn + 6. If n is the definitions of Art. 75, we see that the position of the bovnidary line is alone sufficient to determine the trigonohence all coterminal angles have metrical ratios of the angle From ; the, same trigonometrical For instance, sin (n ratios. . 360 + 45°) = sin 45° = s/2' cos ( 27117 + http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 90/467 8/12/2019 Elementary Trigonometry by Hall and Knight TRIGONOMETRICAL RATIOS OF ANY ANGLE. VIII.] 69 Draw the boundary lines of the angles 780°, - 1.^0°, - 400°, and in each case state which of the trigonometrical functions Example. are negative. (1) Since 780 = (2 x 360) + 60, the radius make two complete revolutions and then turn through 60°. Thus the bound- vector has to ary line is in the first quadrant, so that the functions are positive. (2) Here the radius vector has all to revolve The through 130° in the negative direction. quadrant, third the in line thus is boundary MP OM are negative, the sine, and are negative. secant and cosecant, cosine, and since (3) Since -400= - (360 + 40), the radius vector has to make one complete revolution in the negative direction and then turn through 40°. The boundary line is thus in the fourth quadrant, and since tangent, cosecant, MP is negative, the sine, and cotangent are negative. EXAMPLES. the quadrant in whicli describing the following angles State 1. 135° 135°. 2. 265°. 3. VIII. the a. radius vector -315°. 4. llTi 8. 6 3 3 after -120°. IOtt 5. lies 4 • For each of the following angles state which of the three pi'incipal trigonometrical functions are positive. 9. 12. 15. 470°. 10. 330°. 11. 575°. -230°. 13. -620°. 14. -1200°. iiT ie ^•^'^ in http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight l*^ 91/467 , 8/12/2019 Elementary Trigonometry by Hall and Knight KLEMENTA RY TRIOOXOMETR Y. 70 [chap. each of the following cases write down the smallest positive coterminal angle, and the value of the expression. In 18. sin 420°. 19. cos 390°. 21. sec 405°. 22. cosec 24. cot-^^. 25. sec 4 ( - 3.30°). 20, tan (-315°). 23. cosec 4.380°. 26. tan 257r 3 ' - 3 Since the definitions of the functions given in Art. 75 are applicable to angles of any magnitude, positive or negative, it follows that all relations derived from these definitions must 79. be true universally. Thus we shall find that the formulae given in Art. 29 hold in sin J. X cosec J = 1, tan A J cos A cos A all x sec sin = -. sin- cases ^-1 = 1, , , , cot that ; ^4 = tan A sni A vl fundamental is, xcot^ = l ; cos . - , ; A + GOH A = l, 1 + tan '^ A = sec^ A 1 + cot^ A = cosec- A. be useful practice for the student to test the truth of these formulae for different positions of tlic bounvlary line of the We shall give one illustration. angle A. It will the radius vector revolve Let its initial position OX till it 80, from has traced out an angled and come into the position OP indicated in the perpendicular to Draw figure. J'. A' FM In the right-angled triangle MF^-\-OM^=OF^ OMT, (1). Divide each term by OP'-; thus or) that is. '^['opj sin- Y' ' A 4-coa- A — http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight ]. 92/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight TUIGOXOMETRICAL RATIOS OF ANY Vm.] ])ividc cacli term of (1) by OJ/-; thus '^ XOMJ that tan- is, Divide each term of (1) 71 AN(iI,K. ^1 \0M) ' + 1 = sec- A. by MP^ ; thus fOMy _ f OP '^[jUPj that i s, I ~\MF, + cot -' A = cosec- A thus appears that the truth of these i-elatious depends the statement OP'^=MF'^ + OM^ in the right-angled triangle OJ/P, and this will be the case in whatever quadrant It only OP on lies. Note. - 0.1/- is although the line positive, OM in the figure is negative. In the statement cos J ^s/l — sinÂ, either the positive The sign or the negative sign may be placed before the radical. of the radical hitherto has always been taken positively, because we have restricted ourselves to the consideration of acute angles. It will sometimes be necessary to examine which sign must be taken before the radical in any particular case. 81. Example cos 126° 53' Given 1. = - ^ find , sin 126^ 53' and cot 126° 53'. Since sin- A 1 + cos'-A = sin Denote 126 ^ 53' by A ; magnitude, we have for angles of A= ± A^ 1 .-. A is . positive. sinl26°53'=+^l-^= + results in the following shewn A .( lies Hence the in the sign + before the radical -— The same cos^ then the boundary line of second quadrant, and therefore sin must be placed - any ^g = |; sss.=(-?)-a)=-?may example. also be obtained The by the metliod used appropriate signs of the lines arc in the figure. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 93/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 72 Example 2. The boundary tan If A= line of 15 — - , find sin VIII. and cos A .4 8 A [chap. will lie either in the aecond or in the fourth quadrant, as OP In either position, or OP'. the radius vector= ^/(15) 4-(8)- = ^289 = 17. 15 Hence sinA'OP^Y^, and sin XOP' XOP= cos 15 COB = 17 , XOP' = 17 Thus corresponding to tan^, there are two values of sin A and two values of cos J. however it is known in which quadrant the boundary line of A lies, sin .4 and cos .4 have each a single value. If EXAMPLES. VIII. b. 1. Given sin 120° 2. Given tau 3. Find cos 240°, given that tan 240° = v'3. 4. If .1 5. li 6. I 7. < 8. 9. r = ^ 135°= - = 202° 37' .1 = 143° 8' .t = 21f)° 52' liven sec and 1, 1 sin A A= and cosyl 2, It'cosvl=--, , v'2 fiiul sin find cos ^1 4 ^ - find tan find cot , and cot and tau 1|, find sec^l = - find sin = - -7-- , 13 27r - 20°. find sin 135°. and cosec 2t Given sin -— 4 find tan , .1 and ^1, .1. sin A. 2Tr and cot~- —4 . and sec -r. 4 A and tanyl. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 94/467 8/12/2019 Elementary Trigonometry by Hall and Knight CHAPTER IX. VARIATIONS OF THE TRIGONOMETKICAL FUNCTIONS. CAREFUL perusal of the following remarks will render the explanations which follow more easily intelligible. A 82. Consider the fraction ~ in which the numerator a has a certain X fixed value and the denominator x is a quantity subject to change; then it is clear that the smaller x becomes the larger does the value of the fraction - become. X ^=10«, _1-=1000«, 10 1000 By making fraction ~ =1000U00(>J. 10000000 the denominator x sufficiently small the value of the - can be made approaches to the value oi- For instance 0, we as large as please; that is, as x the fraction - becomes infinitely great. The symbol cc is used to express a. fiuantity infinitely great, more shortly infinity, and the above statement is concisely written ivhen Again, becomes that if jr is X = 0, the limit of ' - X = cc . a quantity which gradually increases and tinally infinitely large, the fraction - becomes infinitely small is, ivheiix — 'X),' the limit of -^ http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight X —0. 95/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 74 [cHAP. If ?/ is a function of .v, and if when .v Definition. approaclies nearer and nearer to the fixed quantity a, the vakie of y approaches nearer and nearer to the fixed quantity b and can be made to differ from it by as Httle as we please, then h is 83. called the limiting value or the limit of y when x = a. Trigonometrical Functions of 84. 0°. Let A'(^P be an angle traced out by a radius vector OJ' of fixed length. MX O Draw (LV /'J/ pcrpendicidar to s,m.POM = we ; then ^p POM . be gradually decreasing, will also gradually decrease, and if OP ultimately come the angle vanishes and MP=0. into coincidence with f I svippose the angle MP OM Again, can OP POM=-^ becomes coincident 1 lencc POM sin 0° Hence cos when the angle POM vanishes OM. .xo = OM = 1. , -^^-jz. OM PC J/ vanishes, tan 0° Ami = wf,= 0- bvit : witli Also when the angle to ô cosecO = 7-i, = 0. OM 1 1 = °°j = •^ô=n sin = secO , cot U cos , 1 ô =T = 1 1 1 =^ =00. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 96/467 8/12/2019 Elementary Trigonometry by Hall and Knight VARIATIONS OP THE FUNCTIONS. IX.] Trigonometrical Functions of 90° or h 85. XOP Let • be an angle traced out by a radins vector of fixed length. M X O M Draw to PM perpendicular to and ^.V, OV perpendicular OX. By deiinition, OM MP . sin POM^^-p ^^^ p„„ -^ POM= cos , MP POM= -^ „^,,, , , tan . we suppose the angle POM to be gradually increasing, When OP comes will gradually increase and OM decrease. If MP into coincidence with and OJ/ OY vanislies, while MP becomes equal sm 90^ to equal to 90°, OP. OP o Hence POM becomes the angle = ^=1; cos 90°=-^yp = 0; MP OP =00 ... tan90=^=— ^ And cot 90° = = tan 90° sec 90° cosec 90° il. K. li. = = 1 cos 90° 1 sin 90° -^ = 0; 00 _ ~ 1 O' = 1. T. y http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 97/467 8/12/2019 Elementary Trigonometry by Hall and Knight 76 ELEMENTARY TRIGONOMETRY. 86. To [chap. and magnitude of sin A trace the chan;jes in sign as A increases from 0° to 360°. Let XX' and YY' be two straight lines intersecting at right angles in 0. With and any radius centre OP describe a circle, and suppose the angle A to be traced out by the revolution of ()P through the four quadrants starting from OX. Draw /M/ perj)endicular to 0,r A= , . sin and let <>P = /'; then MP , r does not alter in sign or magnitude, we have only to as P moves I'oinid the circle. consider the changes of and since r MP When La .1=0°, the first quadrant., .•. When Til .1/P=0, .4 -sin A MP is .*. I .sin A ,1=1 80°, is sinO° =-= 0. r is positive positive = 90°, MP=r, the second qiiadrant, ^V) lei and .sin 90° positive positive MP = 0, and increasing. and MP is and increasing; = - = 1. and decreasing and decreasing. ai id http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight si n 1 80° = - = 0. 98/467 8/12/2019 Elementary Trigonometry by Hall and Knight A'ARIATIONS OF THE FUNCTIONS. IX.] In .•. When A MP is quadrant, the tldrd is negative and increasing. MP is equal to sm . sin A is MP r, but is negative; hence r 270° In the fourth quadrant, • negative and increasing; sinA =270°, . 77 1. negative and decreasing; is negative and decreasing. - Wh^ J The 87. MP = 0, =360°, sin 300° and = = results of the previous article are concisely shewn in the following diagram =1 sin go A sin A sin positive positive and decreasing and increasing sin iSo =o sin A negative siti A negative and increasing and decreasing sin 270=—l 88. We leave as an exercise to the student the investigation of the changes in sign and magnitude of cos A as ^-1 increases from 0° to 360°. The following diagram exhibits these changes. cos cos COS. I So A and cos negative —o cos A A negative cos decreasing and cos positive and decreasing increasin A and g() cos o- 1 positive increasing 2TU = o http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 99/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELKMENTARY TRIGONOMETRY. 78 A and magnitude of tan A To trace tlm changes in sign increases from 0° to 360°. 89. With the figure of Ai't. 86, tan therefore depend on those of When A = 0°, i/P= 0, MP [OHAP. MP A = ^„ , and its changes as will and OM. 6>J/= r; .-. tan 0° = - = 0. ?• Jn the first quadrant, MP is positive OM is positive .*. tan ^ is A = 90°, MP = In second quadrant, r, increasing, and decreasing positive W lien the and 0M= and ; ; increasing. . •. tan 90° = - = cc. MP is positive and decreasing, OM is negative and increasing .•. When In .1 = 180°, is negative and decreasing. MP = 0; .-. tan 180° = 0. the th Ird quadrant, MP is OM is .•. When In tan^ tan .4 is negative and increasing, negative and decreasing positive .4=270°, 07l/=0; the fourth .'. .-. and increasing. tan270°==aD. quadrant, MP is negative and decreasing, OM positive tan When A = 360°, is .4 is and increasing; negative and decreasing. MP = ; • . . tan 360° = 0. Note. When the numerator of a fraction changes continually from a small positive to a small negative quantity the fraction changes sign by passuig through the value 0. When the denominator changes continually from a small positive to a small negative quantity the fraction changes sign by passing through the value oo For instance, as A passes through the value 90°, OM changes from a . small positive to a small negative quantity, hence -rp- changes sign by passing through the value tan A , 0, changes sign by passing through the value http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight while oo , that PM -jr^,, is cos/f, that is . 100/467 . 8/12/2019 . . Elementary Trigonometry by Hall and Knight - IX.] VARIATIONS The 90. diagram of Art. results THE I'UNCTIOXS. Ol' are 8 ) shewn 79 in the following : go tati negative A tan — o:) tan A positive and increasing and decreasing tan o=-o tant8o~o Ian A and tan A negative positive and iticreasing tan 2-0 The student variations in sign —i:x> now have no will decreasing difficulty in tracing the and magnitude of the other functions. In Arts. 86 and 89 we have seen that the variations of the trigonometrical functions of the angle XOF depend on the position of P as P moves round the circumference of the On this account the trigonometrical functions of an angle circle. This name is one that we shall are called circular functions. 91. use frequently. EXAMPLES. IX. a. Trace the changes in sign and magnitude of ^ ^ V 1. cot 2. cosec 3. cos 4. tail 5. sec d, .4, 0° between 6, between and 360^ and tt. between v and 27r. A, between — 90° and - 6, between - and :iHf, — Find the value of 6. cos 0° sin2 270° 7. 3 sin 0° sec 8. 2 sec2 n 9. i. tan TT TT 1 cos cos - 80° 2 cos 180° tan 45°. + 2 cosec +3 sin^ -- — + sec „ Stt , ztt — 90° - cos 360°. - cosec Stt cosec -5- http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 101/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TKIGONOMETKV. 79, [CHAI'. Graphs of the Trigonometrical Functions. The 91a. results of Arts. 86 — 90 may l)e illustrated liy uieans of graphs. Put y = siii,r, ami timl the values of sin corresponding to values of .v dilfering by 30°. Graph From .. . .<•. a Table of sines of y we have 0° http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 102/467 8/12/2019 l.\.] Elementary Trigonometry by Hall and Knight GKAPU.S Ub' THK TKXUUNOMKTRICAL FUNCTIONS. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 79jj 103/467 8/12/2019 Elementary Trigonometry by Hall and Knight KLKMENTAKY TRIGONOMKTKY. ra, We [CHAI'. the details of the graph of an exercise for the student. It should be drawn from the same series of angles, and with the same iniits as in the It is given on a small scale in the adjoining graph of sin.i'. 91b. cos X as G-raph of cos .'. le;i\e tigure. Y +1 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 104/467 8/12/2019 IX..] Y Elementary Trigonometry by Hall and Knight GKAl'HS OF THK TKIGOHOMETRICAL FUNCTIONS. J] 79jj ' 1 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 105/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARV TRIGONOMETRY. 79., EXAMPLES. 1. IX. [cHAT. b. with the same axes and units, the grai)hs of sin l)ra\v, ./• I.' scale twice as large as that in Art. on a and COS. 91^4^. [This diagram will be required for Ex. 2. Draw the graph of sin 30°, 15 , 5°, .v 45^ 9.] from the following values of 60°, 85°, 75°, .v : 90°. inch horizontally to represent 25°, and 2 inches vertically to represent unity.] [Take 3. 1 angle whose sine 4. figure of Ex. 2 find the value of sin 37 , ;uid the From the to the nearest degree. is '8, Find from tlie values 0°, 10 Tables the value of cos V Draw a curve on a from inci-eases 5. cos 25° 6. From and cos 45°. lias tlie 60°. how large scale shewing .v x cos varies as :v 4, values of approximate find Verify by means of the Tables. Solve graphically the following equations, giving, to the all the solutions less than 360°. 15sin2^-16shi(9 + 4 = 0; cos2(9-l-7 cos(9 + -72=0. (ii) Draw the graph of cot.r, using the following values of 15°, 30°, [Take 1 45°, 60°, figure of Ex. Verify l)y Fr<MU the graphs in Ex. 7 find means J, approximate + cos.v=0; of of the Tables. .I-. solve the following equations graphiwilly sin.f ^•ahles deduce the graph of sin .r+ cos (i) 180°. ... inch horizontally to represent 15°, and inch vertically to represent unity.] From the Hence 105°, 90°, 75°, .v. 1 cot 48° and cot 59°. 9. 50°, when 0° to 60°. (i) 8. 40^ 30°, the figure of Ex. nearest degree, 7. 20°, .f (ii) http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight sin j.' ; + cos.i'— I. 106/467 8/12/2019 Elementary Trigonometry by Hall and Knight THE IX.] DKKlNlTiOlNs. OJ.L) Note on the old dejinitions of the Triijonometrical Functions. Formerly, Mathematicians considered the trigonometrical functions with reference to the arc of a given circle, and did not regard them as ratios but as the lengths of certain straight lines drawn in relation to this arc. Let OA and OB be two radii of a circle at right angles, and let P be any point on the circumference. Draw and PN perpendicular to OA and OB respectively, and let the tangents at A and B meet OP produced in T and t respectively. PM The Un^s PM, AT, OT, AM were named respectively the sine, tangent, secant, versed-sine of the arc AP, and PN, Bt, Ot, BN, which are the sine, tangent, secant, versed-sine of the complementary arc BP, were named respectively the cosine, cotangent, cosecant, coversed-sine of the arc A P. As thus defined each trigonometrical function of the arc centre of the circle, AT —---ta.nPOA; that OA The is, .-12'= 91 =:sec BOP -cosecPOA; and is which it subtends multiplied by the radius. Thus to the corresponding function of the angle, OB 0^ that x tan is, PO^ Ot = OB equal at the ; xcosec POA. values of the functions of the arc therefore depended on the length of the radius of the circle as well as on the angle subtended by the arc at the centre of the circle, so that in Tables of the functions it was necessary The names them now to state the magnitude of the radius. and the abbreviations for in use were introduced by different Mathematicians chiefly towards the end of the sixteenth and during the seventeenth century, but were not generally employed until their re-introduction of the trigonometrical functions The development of the science of Trigonometry may be considered to date from the publication in 1748 of Euler's Introductio in analysin Infinitornm. by Euler. The some interesting information regarding the Trigonometry ia Ball's Short Ilistorij of Mathematics. reader will find l)iogrcss of http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 107/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight ELKMENTARY TRIGONOMETRY. 80 [CHAP. MISCELLANEOUS EXAMPLES. Draw the boundary 1. Lxpial to 3 -- and , lines of the angles C. whose tangent is find the cosine of these angles. 4: Shew that 2. cos A (2 sec A + A) tan 3. Given C=90°, 6 = 4. If secJ.= — 25 V. A-2 tan A) = 2 cos ^1 10-5, c and ' 7 (sec A = 2\, lies - 3 tan A. solve the triangle. between 180° and 270 , find cot A. The Bombay N. find its distance from the equator, taking the diameter of the earth to be 7920 miles. 5. latitude of is 19° : From the top of a cliff 200 ft. high, the angles of depression of two boats due east of the observer are 34° 30' and 18° 40' find their distance apart, given 6. : cot 34° 30' 7. If A lies value of 2 cot 8. = 1 -455, cot 1 8° 40' =2 96. between 180° and 270°, and 3 tan A— 5 cos Find, correct .4 = 4, find tlie A + sin A to three decimal places, the radius of a which an arc 15 inches long subtends at the centre an angle of 71° 36' 3-6 . circle in 9. Shew that tan3 d cot^ d l+tan2^ 10. The angle 1 + cot^^ 1-2 sin^ d cos^ sin (9 cos ^ of elevation of the top of a tower is 68° 11', and a flagstaff 24 ft. high on the summit of the tower subtends Find the height of the an angle of 2° 10' at the observer's eye. tower, given tan 70° 21' = 2-8, cot 68° http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 1 1' = -4. 108/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 109/467 8/12/2019 Elementary Trigonometry by Hall and Knight CHAPTER X. niHCULAU FUNCTIONS OF CERTAIN ALLIED ANGLES. Circular Functions of 180° - A. 92. Take any straight line XOX', and let a radius vec- , -v, OX revolve tor starting from until it has traced the angle A^ taking up the position OP. gol^ ^ ''-^^ ~~--~^^^^^ MX O x'M' OX revolve through Again, let the radius vector starting from 180° into the position OX' and then hack again through an Thus XOP' is the angle A taking up the final position OP. -J. angle 180° From then b}' P and P' draw Euc. i, PM 26 the triangles and PM' perpendicular to XX' ; OPM and 0PM' are geometrically equal. By definition, sin (180°-- hut .1/7 is equal to .-. Again, and OM' is . equal to .,0^0 ( 1 80° OM in .lono cos (180 lan(]80 magnitude and sin(180 -.-l) cos .-. Also MP in M'P J ) = -^-; - .1 ) is of the same sign ; MP = -^ = sni.l. = -^- \ magnitude, but -OM OM is of opposite sign , ,^ = -^p-=--^=-cos.4. -^) M'P MP ,^ -A)=-^,== 1^^= http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight MP ^ 03/= -^• ^, 110/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight CIRCULAR FUNCTIONS OP CERTAIN ALLIED ANGLES. 83 In the last article, for the sake of simplicity we have supposed the angle A to be less than a right angle, but all the formulse of this chapter may be shewn to be true for angles of any magnitude. A general proof of one case is given in Art. 102, 93. and the same method may be applied 94. the If angles to all the other cases. expressed are measure, radian in the formuhe of Art. 92 become sin (tt Example - ^) = siu 6, cos (tt — ^) = — cos 6, tan (it — 6)= — tan 6. Find the sine and cosine of 1. sin 120° = sin (180° - 60°) = sin 120°. 60° ='^ • co8l20° = cos(180°-60°)= -cos 60°= --. Example Find the cosine and cotangent of ^^ 2. . cos^=cos(^x-gj=-cos^-=--^. cot -r 6 95. Definition. = cot ( 7r r 1 0/ \ When the = - COt 6 sum = - v/3. ^ of two angles is equal to two right angles each is said to be the supplement of the other and the angles are said to be supplementary. Thus if A is any angle its supplement is 180° - .1 96. The results of Art. 92 are so important in a later part of the subject that repeat them it is desirable to emphasize them. in a verbal form therefore : the sin es of supj> lementary angles axfusqual.in are of the same sign We magnitude and ; of supplementary angles are equal in magnitude hut are of opposite sign; the cosines the tangents of supplementary angles are equal in magnitude but are of opposite sign. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 111/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 84 [f'HAP. Circular Functions of 180°+ A. 97. Take any straight line XOX' and let a radius vector starting revolve until it has traced the angle A^ taking up from OX the position OP. Again, let the radius vec- OX revolve tor starting from 180° into the position through OX, and position then further through an angle J, taking up the OP. Thus From P and then OP and the triangles By XOF is the angle 180° and P'M' perpendicular to P' draw XX' same straight line, and by Euc. OP are in thePM 0PM and OPM' are geometrically equal. ; 26 I. definition, sin(180°+yl): M'P and final + J. is .sin (180° is equal to OP Also op'' + .4) = of opposite sign ; sin A. ^; magnitude but is of oppo.site sign ; OM cos(180 +.4) = --^^^= --^ = - cos A -MP MP 31'P ^• tan (180° +^) = ^ OM'' -OM ^ ,^=^'=**'^ ^Y' , .-. is MP _ MP + .4) OM in OP magnitude but cos (180° Again, and OM' MP in equal to M'P '' oô .s -OM . Expressed in radian measure, the above formulae are written sin (tt + ^) = - sin cos ^, In these results we (tt + <9) = - cos ^, may draw tan (rr + ^) = tan ^. especial attention to the fact that an angle may be increased or diminished by two right angles as often as we please without altering the value of the tangent. Example. Find the value of cot cot 210° = cot (180° 210°. + 30°) -: cot http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 30° = ^S. 112/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight CIRCnLAR PDNCTIONS OP CERTAIN ALLIRP ANGLES. X.] 85 Circular Functions of 90* + A. 98. Take any straight line XOA'', and let a radius vector starting revolve until it has traced the angle A, taking up from OX the position OF. Again, let the radius OX starting from through 90° into the tor vec- revolve position OV, and then further through X' an angle A, taking up the final position Thus XOP' is the angle 90° + .1. and P' draw PM and PM' perpendicular or. From P then z Bv Euc. equal ; M X O M' I. M'FO= P'0Y=A = l 26, the triangles i to XX' POM. OPi/and OPM' are geometrically hence M'P' is equal to OM in magnitude and is of the and OM' is equal to MP magnitude but is of opposite sign. By in same sign, <lefinition, sm(90 +^)=—^=— = cosJ; cos (90 + .4): M'P ,^.„ tan(90+.^^=— ,. , OP OM =_;^^ Expressed in radian measure sinf^ + ^j Example = cos(9, 1. cos^l + H. K. tlie ^W = sin (90° + A 0P~ OM = MP -^ cot .1 above formuliie beconif -sin^, Find the value of sin sin 120° sin '' OP . MP -MP DM' tan (| + (9)= cot(9. 120^. 30°) -cos 30° = '^. K. T. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 113/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 80 Example Fiud the values of tan (270° + A) and cos (-^ + e\. 2. tan (270° + J) = tan (-^+^| = cos (180° cos(7r + 90° + ^) = tan + ^+^) = Take any straight OX line (90° - cos +^ = ) -+ ( ^ ) - cot ^ = ; -iin ^. -A. Circular Functions of 99. [chap. and let a radius vector starting from (^V revolve initil it has traced the angle A, taking up the position OP. Again, ing from let the radius vector start- OX revolve in the opposite imtil it has traced the taking up the position OP'. angle A, Join then MP' is equal to direction Pr MP ; in magnitude, and the angles at are right angles. Bv [Euc. I. M 4.] definition, Hm{-A) = ,, , cos(-y1)= , OF ~ OP OM = OM = ^ ^ sin ' cos -MP MP' ,- , MP _ -MP A Example. altering the value of ; ; tan It is especially worth}' of notice that v:e of an angle without A A. may change the sign its cosine. Fiud the values of cosec ( - 210°) and cos {A - 270°). coaec(-210°)= -coscc210°= -cosec (180° + 30°) cos (A - 270°) = cos ^ (270° -cos -A) = cos (90° (180° -.4)= -sin http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight + = cosec 30° = 2. 90° - A) .4. 114/467 8/12/2019 X.] Elementary Trigonometry by Hall and Knight C'lRCULAR FUNCTIONS OP CERTAIN ATiLIED ANGLES. 87 /(^) denotes a function of A which is unaltered in magnitude and sign when —A is written for A, then/(^) is said to be an even function of A. In this case/( — ^)=/(^). 100. If for A, the sign of f{A) is changed If when -A is written while the magnitude remains unaltered, f{A) is said to be an odd function of ^, and in this case/( - A)= — f{A). From the last article it will cos ^-1 and sec A be seen that arc even functions of A, A are EXAMPLES. X. sin A, cosec J, tan A, cot odd functions of A. a. Find the numerical value of 1. 003 135 . 2. sin 150°. 3. tan 240°. 4. cosec 225°. 5. sin (-120°). 6. cot (-135°). 7. cot 315°. 8. cos (-240°). 9. sec (-300°). 10. 13. tan 11, . 4 rosec^-iy 14. sm —- 12. . o A — . »> oo.s(-^V Express as functions of .sec 15. cot^-^-^' : 16. cos (270° + ^). 17. cot (270°-. ). 18. sin (^-90°). 19. sec (J. -180°). 20. sin (270° -J). 21. cot 24. sec Express as functions of d 22. sin (^ - 1) • 23. (.-1 - 90°). : tan {6 - n). (^ - ^) • Express in the simplest form 25. tan (180° 26. cos (90° 27. sec (180° + ^) sin (90° + .1) sec (90° - A). + vl) + sin (180° - J) - sin (180° +A) - sin - A). ( + J)sec(180'-yl) + cot (90° + http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight .-I) tan (180° + i4). 115/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 88 [chap. In Art. 38 we have established the relations which subsist between the trigonometrical ratios of 90° — A and those AVe shall now give a general of A, when A is an acute angle. proof which is applicable whatever be the magnitude of A. 101. -A Circular Functions of 90° 102. for any value of A. X'M Let a radius vector starting ^froni OX revolve until it has traced the angle A, taking up the 'position OP in each of the two figures. OX revolve through the radius vector starting from )0° into the position and then back again through an angle,-J, in each of the two figures. taking up the final position Again, Draw let OF PM OF and P'M' perpendicular to XX'; then whatever the value of A, it will be found that z OFM'= l POM, and OM'P' are geometrically equal, so that the triangles equal to M'P\ in magnitude. equal to 0M\ and having be OMP MP OM is above XX', F is to the right of YY', P is below XX', P' is to the left of YY'. P' is above XX', P is to the right of YY', and when P' is below .r.V, P is to the left of YY'. When P and when When Hence MP is equal the same sign as OM' to OM' in magnitude and is always of ; and M'P' is equal to same sign as OM. OM in magnitude and http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight is always of the 116/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight l'[RCULAU KIFNOTIONS OF CKRTAIN ALJ>XED ANGLES. X.] By A 89 detinitioii, ^- = -^ = siu A COS (90 - .4 ) = tan (90 -^) = ^^ = ; ^^ = cot4. may general method similar to the above be applied to all the other cases of this chapter. Circular Functions of 103. If n any is integer, n . 360 + A. n 360° represents n complete . revolutions of the radius vector, and thei-efore the boundary line of the angle n. 360° + J. is coincident with that of ^4. The value of each function of the angle n 360° + ^ is thus the same as the value of the . corresponding function of A both in magnitude and in sign. coterminal angles are equal, thei-e is a recurrence of the values of the functions each time the boundary line completes its revolution and comes round into its original position. This is otherwise expressed by saying that the circular functions are periodic, and 360° is said to be the amplitude Since the functions of 104. all of the period. In radian measure, the amplitude of the period Note. 7r is radians. 105. If n incident tangent and cotangent the amplitude of half that of the other circular functions, being ISO ^ or In the case of the period is 27r. tlie [Art. 97.] Circular Functions of n 360° - A. . is any with boundary integer, the that of -A. line of The value m.360°-.1 each of is function co- of M.3(i0°-.l is thus the same as the value of the corresponding hence function of ~ A both in magnitude and in sign ; sin (« c<js (n . . 360° - A) = sin 360° - A tan (m 360° . . I ( - J ) = - sin ) = cos — A) — cos ) = tan {- { A)=^ A A ; ; — tan ^1 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 117/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 90 [CHAP. We can always express the functions of any angle in terms of the functions of some positive acute angle. In the arrangement of the work it is advisable to follow a uniform 106. plan. If the angle is negative, use the relations connecting the (1) functions of — ^ and Thus sin cos ( [Art. 99.] ,1. - ( - = - sin 30° = - ^ 30°) 845°) ; = cos 845°. (2) If the angle is greater than 360°, by taking oft' multiple.s of 360° the angle may be replaced by a coterminal angle less than 360°. [Art. 103.] Thus tan 735° = tan (2 x 360° + 15°) = tan If the angle is stiU greater (3) than 180°, use the relations connecting the fimctions of 180° + -4 and A. Thus cot 585° 15°. [Art. 97.] = cot (360° + 225°) = cot 225° = cot(180° + 45°)=cot45° = l. If the angle is still greater than 90°, use the relations (4) connecting the functions of 180°- J. and A. [Art. 92.] cos675° = cos(360° + 315°) = cos315° = cos (180° + 135°) = - cos 135° Thus = - cos (180° - 45°) = cos 45° = -t;^ Example. Express sin (- IIQC^), tan 1000°, cos (- . 980*^) as functions of positive acute angles. sin ( - 1190°) tan 1000° = - sin 1190° = = - sin (180° - - sin 70°) ( - 980 ) x 360° - sin + 110 =) = - sin 110° 70°. = tan (2 x 860° + 280°) = tan 280° = tan (180° + 100°) = tan 100° = tan (180° - 80°) = cos = (3 = cos 980° = cos ( - tan 2 x 360° = cos (180° + 80°) = 80°. + 260°) = cos 260° - cos 80°. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 118/467 8/12/2019 Elementary Trigonometry by Hall and Knight CIRCULAR FUNCTIONS OP CERTAIN ALLIED ANGLES. X.] 91 From the investigations of this chapter we see that the number of angles which have the same circular function is unlimited. Thus if tan^ = l, d may be any one of the angles 107. coterminal with 45° or 225°. 6 Draw Example. write down all the boundary lines of A when sin A , aud the angles numerically less than 360° which satisfy the equation. Since sin 60° = ''^ is , if we draw one position of the boundary OP making /. A'OP = 60'=, then OL' line. Again, sin 60= ^ sin (180° - 60°) = sin 120°, so that another position of the boundary line will be found by making .YOP' = 120°. There line will be no position of the boundary in the third or fourth quadrant, since in these quadrants the sine is negative. Thus in one complete revolution OP and OP' are the only two positions of the boundary line of the angle A. Hence the positive angles are 60° and 120° and the negative angles are - (360° is, -240° and -300°. 120°) EXAMPLES. X. and - (360' 60°); that b. Find the numerical value of 1. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 119/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight ELKMKNTARY TRIGUNOMETUV. 92 l 'iiid the angles numerically less than 360 all the equations : cos^ = 21. = 23. tan If is less ^-1 (9 - ^/.3. than .sin^=-i. 24. cot 90^, 26. tan(270° 27. cos 28. Prove that (.4 .-1 1 prove geometrically + ^)=-cot.l. -90°) = sin J. that - A) 1 80° - ^) cot (90° tan (180° + ^) tan (90° +.1) ( = - + tan (180 - .1) + cot (90' + .1) = tan (^360 - .1). Shew sin (9 (^-180°)= -sec ^. sec 29. 22. ^. 25. tan which satisfy cos (360° -A ) - sin(-^) ' = sin^(. Express in the simplest form - sin(180° „. + J) , cos J. sin(90° + ^)' cos(-J) + ^) cos(90° + ^)* + ^) sec - A) tan (180° - A 80° + ^ cot (90° -A)' sec (360° + ^ sin cos (90° ( ( 1 ). 33. + .1) cot^ cosec(180°- J) sec(180° 32 tan (90° sin (-.-I) Prove that sin (J + d\ ) cos (n - 0) cot ( ^+ 19 = sin(|-^)sin(^-^)cot(| + ^) 34. When a= Utt 4 sin- a , find the numerical value of — cos- a + 2 tan a http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight — sec- o. 120/467 8/12/2019 Elementary Trigonometry by Hall and Knight CHAPTER *^ XI. FUNCTIONS OF COMPOUND [If preferred, Chapters XIII, before 108. When Chapters XI XV mat/ (>< (nice it and XII.\ made up by the algebraical t<iini of thns A + A', called a compound angle compound angles. an angle two or more angles it is A - i?, and A + B — C are XI\\ AN(J1.KS. is ; Hitherto we have only discussed the properties of the In the present functions of single angles, such as A, B, a, 6. chapter we shall prove some fundamental properties relating to We shall begin by finding the functions of compound angles. 109. — expressions for the sine, cosine, and tangent oi A-\-B and in terms of the fiuictions of A and B. may A B be useful to caution the student against the prevalent is equal to the sum mistake of supposing that a function of yl of the corresponding functions of A and B, and a function of A - B to the difference of the corresponding functions. It +5 +B) is not equal to sin A +sin B, cos (A - B) is not equal to cos A - cos B. Tlius sin {A and A numerical instance will illustrate 'J'luis if .1 = 60', 5 = 30°, cos {A so that then this. A +jB = 90°, + B) = cos 90° = ; 2 Hence cos (.1 + B) is not equal to cos In like manner, sin [A tliat is, Similaii} -\-A) is J +cos 2 B. not equal to sin A +siu sin 2.1 is not e(jual to 2 sin A. tanlLI is not equal to .'Han A. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight A ; 121/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 94 110. 7h prove the formulae +B) = sin A sin (J cos 5+ cos A sin B, 5 - sin -4 iLOM=A,sind lMON=B\ then +B) — cos A cos {A Let [CHAf. cos sin z if. ZOiA -=.l In ON^ the boundary line of the compound angle point P, and draw respectively ; also A + 5, + ^'. take any and PR perpendicular to OL and OM PQ ^*S' and RT perpendicular to OL and P(^ draw respectively. By definition, sin(^+5) = PQ ^ RS+PT = --::^^^ = PT RiS -^+^ BS OR PT PR ^ OR' OP^ PR' OP = sin A cos 5 + cos TPR .sin B. . But L TPR = m°.'. Also sin (vi „, L TRP= + 5) = sin J. OQ ,, = ^-p = cos(J+5) , L TRO^ iROSÂ cos 5 + cos ^4 OS-TR OS _^^^ = sin 5. TR ^-^ _0S OR _ TR PR ~ OR OP PR OP ' = cos A = cos A . cos /i- sin cos TPR B - sin A http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight . sin U sin B. 122/467 8/12/2019 Elementary Trigonometry by Hall and Knight 95 FUNCTIONS OF COMPOUND ANGLES. XI.] To prove 111. sin the formulce {A-B) = sin Aco&B- cos A sin cos {A Let L -B) = cos LOJf^ A and z , In OiV, <Ae boundary sin B, A sin i>. AcohB + i/OiV - iJ line of the ; then i LON= A - B. compound angle A ~B, take anv perpendicular to OX and OM Lnd draw PQ and PR to OL and also draw RS and RT perpendicular ?espectilely point P, ; QP respectively. By definition, sm{A-B) = ^ = —0P~ PQ RS RS^ PT ' 'OP OP R&-PT qR_RT PR PR OP = sin ^ cos B - cos TPR - OR -OP . But / sin {A ^ Also sin B. TRP= i MRT= l M0L=A - B) = sm A cos 5- cos A sin B. TPR = 90''.-. . I coii{A-B) = OQ OS+RT_qS OS OR.RT RP ^— RT ^n-t- OP'^ OP OP ÔR- OP'^ RP'Of = cos .4 cos B + sin TPR . = cos A ; cos £ 4- sin J http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight . sin B sin B. 123/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 96 The expansions of [CHAl*. and cos (^ +^) are fre(lueiitly called the Addition Formulfe. We shall sometimes —5 ^4+5 and ^ lefer to them as the formulae. 112. sin(.4 +j5) In the foregoing geometrical proofs we have supposed that the angles A, B, A + B are all less than a right angle, and 113. that A- B If the angles are not so restricted \^ positive. of the modiiication figures be will required. It is some however unnecessary to consider these cases in detail, as in Chap. XXII. we shall shew by the Method of Projections that the Addition Formulae hold universally. In the meantime the student may as^sume that they are always true. E.raiuplc 1. cos 75° Find the value of cos = cos (45° + 30°) = cos 45° sin But 2^/2 = ^ and sin B~ —5 (A- B) = sin A cos B- cos If sin os 4 -4 A= s/l - sinÂ — a / ..._,. Note. Strictly speaking 4 1 3 12. • A sin B. 163 25 / / 3 cos^=±- sin 45° sin 30° find sin (A - B). , 1 cos£= Vl-siu-^^'Y ami - cos 30° ^/2 2 2 V2' Exaiiqilc 2. 75°. - ~ 5 25 = jgQ 5 ' 12 jgl 33 12 and cosj6'=:± — , so that - B) has four vahies. We shall however suppose that in similar cases only the positive value of the square root is taken. sin (A 114. The To prove that sin {A first = (sin +B) sin {A - B) = sin^ A — sin- B. side A cos ^4- cos = sin'^ A coii^ •=sijr- (1 -I = sin->l - A sin 5)(sin B - cos- A sin- 7^)- (1 sin^ — A cos7i — cos^l sin B) B sin- A) sin- -sin-y/. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 124/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight FUNCTIONS OF COMPOUND ANGLES. XI.] EXAMPLES. XI. 97 a. [7%e examples printed in more prominent type are important, and should he regarded as standard formulas^ Pi'ove that , y 1. .sin (yl + 45°) = -^- (sin (/' 2. cos {A + 45°) = -y^ (cos A - sin J [/ 3. 2 siu (30° 1^ 4. If cos .4 5. If sin A=-, 6. If sec ^=—, *^ .-l+cos A). -A) = CCS A - ^3 sin = |, o cos i? = ? cos j?=-— co.sec^ ). .4 find sin {A , + B) find cos (A = -, 4 and cos {A - +B) and find sec (J sin (A B). - 5^. +B). Prove that / 7. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 125/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 98jk^ Prove the following identities ^ 18. 2 sin {A + 45°) sin 19. 2 cos +a 20. 2 sin 21 ^^ -fcx) 1/ ( 2 (^ + cos ^ cosy cos ) : {A - 45°) = sin^ ( 7 aj cos f| -y + ^ j =cos(a + ^)+sin cos a tan To expand + 5) = „, (.4 ^ ^ (g-^) ^ (a-/3V ^^ cos a cos j3 terms (^4 + 5) sin ( J^ -^B) „; cos(J.+5) ; a. ) tan 115. A - cos^ A. - a = COS^ a - sin^ sin + î^^- ) + cos [CHAP. , = m sin tan ^4 o?i£/ tan B. of 5+ cos A sin 5 p^ — —j-^ „ cos ^ COS ^ - sin ^ sin /i J. cos . j . express this fraction in terms of tangents, divide each term of numerator and denominator by cos A cos B To ; B COS 5 COS^ s.; sm A sm B — sin , , „, .-. ' ^ tanM+i5) = A 5 • COS that A „, ^'^ (^'^+^) IS, we may Example. ' cos B = rruKj-taKîs given in Chap. XXII. i)rove that tan ^^ A • A + tan B tan geometrical proof of this result Similarly, sin A — tan B ^^^-^^^l+t^nAt^nB- Find the value of tan 75°. tan 45° tan ^•^ -^^ (^•'^° + tan 30° + ^Q°) = l-t.an45°tan30° _ (n/3 + 1)(\/3 + 1 ^ 4 + 2^ 3 3-1 2 = 2 + ^/3. ) http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 126/467 ' 8/12/2019 Elementary Trigonometry by Hall and Knight FUNCTIONS OF COMPOUND ANGLES. XlJ To expand cot 116. , , , cot (A To „, +B) — tenns o/cot -\-B) in (.-l — A and cot 7>. -. )—. -. . . fz j5 express this fraction in terms of cotangents, divide each , . . , cot {A „, +B)= cos A cos B sin A sin B „ Similarly, 5 we may prove cot (A ^ 117. , sin -B)= ' = sin {A+B) = sin ..4 To find ,. tan(.4+i^ the 5+1 --jcot jS — cot A —— p .sin cos cos T 5 + cos ^4 sin B) cos C + (cos A cos i5 - .sin A B cos C'+ cos TIN i .4 cos (.1 sin sin 5) sin (J B cos ^ BmiiC-axwA — 1 1 tan (4 +5) + tan 6' bu^6' i_tan(,4-l ,^ , i^ • tan tan ^ +A - tan tan ^4 tan tan B+ 5 - tanC— B tan C— tan B ( tan numej-ator of the above expression .4 T'. +B + C). + i5+C=180°,thentan(^+5+C) = 0; tan sin i? sin ^ + tan B ^ ^ + tan L 1 — tan A tan 5 tan ^ + tan -^ ^ ^^'^ ^ 7? TT 1 — tan A tan B tan .•. . cos f'+ cos {A -^B) sin expansion o/tan ,, 1 If.4 • -\-B+ C). (^1 tan CoR. B A ^ .V. 0-tan{(^+5) (7+ + , — ; cot.-l cot + COS.4 118. B + j5) + C} {(^ = (sin A sin that To find the expansion of J + 5 + C) = sin A - 1 cot A cot ~^5 r-r cot ±1 + cot A = cos^ cosxj sin ( <Q^ COS A COS B — sin A sin ^ + B) ^'^ = —. — —Zj sin{A + B) sm J. cosi5 + cos^ Sin — cos (A term of numerator and denominator by sin sin . + tan ^ + tan must be (^=tan .4 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight ' tan C A ' hence the zero. tan ^tan ('. 127/467 8/12/2019 Elementary Trigonometry by Hall and Knight KT.EMENTARY TRIGONOMETRY. 100 EXAMPLES. \TIie XI. [cHAP. b. examples printed in more prominent type are important, and should he regarded as standard formulce. 'I ./ 1. Find tan *^ 2. If tan {/ 3. If cot .l ^-1 {A+B) when =^ =^ , and , ^ 4. If cot c 5. ^ L/6. , find cot , A-=^, tan />iKAN = tan(45+A) , B = ^, tan ( A B=- - B). find tan (.4 5 7 C^ B = 45°, ^ =^ cot A — -, tan + B) find cot {A and tan {A ^ - B) and tan (A H). + B). 1+tanA 3-_-:^^^^^. ^ tan (45^- A) =j,^:^^;^. cot^ + 1 ./^ -7 Q ^ /»r -\ cot^-1 A ^'^ 9. ./\\. tanl5° = 2-v'3. = 2 + s'3. Find the expansions of cos {A '^ cot 15° 10. + B + C) and sin {A-B+ C). 12. Express tan (^ -i5 -(7) in terms of tan 13. Express cot {A + B + C) in ^4, terms of cot A , tan tan yi, cot B, cot < ('. '. :^ Beginners not unfreqnently find a difficulty in the conand A- B formulte that is, they fail to verso use of the i-ecognise when an expression is merely an expansion belonging to 119. A+B ; one of the standard forms. Example 1. Simplify cos This expression is (a - ^j cos (a + /3) - Example ^ By 2. -j3), /3) sin (a the expansion of the cosine of the therefore equal and Shew that — - 1 + /3). compound (a-;8)} to cos {(a is angle (a + /3) + (a that is, to cos 2a. sin (a - + /3) + — 7_ ;=taud^. - tan A tan 2 Art. 115, the first side is the expansion of tan (A +'2A), and is tlierefore equal to tan HA. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 128/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight CONVERSE USE OF THE ADDITION FORMULAE. XI.] Example Prove that cot 2^ + tan 3. llic nrst side cos 2 A = ———sm 2A _ ~ cos (24 -^) sin 2A cos 4 =cosec 2A. .J —A + sin 2 A sin A A r = cos A cos 2 A cos siu M 101 ^;-, -. sm 24 ; cob A ^'^^ '^ _ sin 24 cos A = -^—^r— = cosec A sm24 '2 Example cos The first Prove that i. 4:0 + sin 48 cos d side = cos (4^ sin ^ = cos 26 cos ^ -6)= cos 3(9 = cos = cos 2^ cos 6 - sin 25 siu EXAMPLES. Pi'ove the following identities '^ cos (.4 +5) sin 3.4 cos 1^3. cos 2a cos a ^' ,/ cos + sin 6. XI. c. 7?) sin 5 = COS. 2a sin a = cos a. - .4 ) - si n .4) cos (30° + .4) + cos (60° - + .4 5. sin (60° - - cos 2a sin 2a sec a cosec a = cos 6. tan(a-/3) + tan/3 I. A = sin 2 A. cos (30° cos (30° (^. + 6) ) 4. ,/ _ sin sin 26 siu : 5+ siu (.4 + A - cos 3.4 1/ 2. <^ y 1. {26 - (30° =^ ) si n (30° - .4) sin (.30° + .4) = 1. + .4 -'1 ) • 3n. , l-tan(a-/3)tan^ ^8. ^ôt(a cot ^ 9 ^ 10. 11. + 3)cota + l^^^^^ a— cot (o 4-/3) tan4^zt^3J^ = tan.4. I + tan 4 J. tan 3 A - cot 2<9 = cosec 26. cot 1+ tan 2(9 tan = sec 2(9. (9 + cot 20 cot ^ 12. 1 13. sin 2^ COR ^ 14. c<,)S H. K. = cosec + cos 2B cot 2^ sin ^ 4a cos a - sin 4a siu a JO. =^ ^. siu 4^ cos fl- cos 4^ siu — cos 3a T. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight cos 2a - sin 3a siu fi. 2(i. 'o 129/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 102 [(HAP. Functions of Multiple Angles. To express sin 120. 2.1 in terms of am A and cos./ . 2.-1 = sin {A-\-A) = sin A cos A + cos .4 sin 2A = 2 sin A cos A. sin that i.s, may have any sin A ; a perfectly general fornuila for the sine of an angle in terms of the sine and cosine of the half Thus if 2 A be replaced by ff, we have angle. Since .1 value, this Sin 6 is = 2 sm - cos 2 Similarly, 4A = 2 sin sin 2A = 4 sin A 121. express cos 2.4 7'o cos 2 A that cos cos tVi cos . 2 2A xi cos 2 A êrms of cos = cos (^4 +A) = cos ^4 is, rr cos 2A = cos^ ^4 - ..4 J. ««(^ sin — sin A sin ^1. A ; sin^ ^4 (1). There are two other useful forms in which cos 2A may be one involving cos A only, the other sin A only. expi-essed, Thus from (1), cos that 2A = cos^ A — {1cos2.4 is, Again, from = (1 cos is, From fonnuho = 2cos2^-l (2) - sin^ 2^ = 1 - and (3), we A) — 2 and • sin'- sin- A ; J (3). 0082.-1 (4), l-cos2J=2sin2 J '...(5). 1- T By division, • (2). obtain by transposition ]+ cos 2.4 =2 ij A) (1), cos 2^4 that cos^ cos 2.4 ,„^ ., ^^ ^^^- i:^^-2A=^^''' Example. From (8), Express cos 4o in terms of cos4a— 1 - 28in22a = l-2 sin a. (4 sin'-'a cos a) — 1 - 8 sin o (1 - sinô) = 1 - 8sin a-l-8sin''a. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 130/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 131/467 . 8/12/2019 . Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 104 EXAMPLES. and should ^ =o be rec/arded as 1. If cos 2. Find cos 2 A when sin 3. If sin 4. If tan ^ = -, find tan 2^. 5. If tan 6 = \, find sin 6. If cos 7. Find tan ly -i ^^^ ^ ' ^^^ -^ 5 . 3 y A=-, 8. ^ ,rt 10. 2 J = -. 5 2(9 ftnd tan ^ a=Ti A when and cos cos 2 J 2 cosec « cosec 16. cot 2A Ti^ cosec - l4-sec(9 — ;r- sec 6 A T-î— 2 cot l+COt2J - „„ „ — COt-A-1 ...oA cot .4 2 ^r-:, 22. A + cosA T^Â^^cot sin A . J tan a 18. 20. cos a. 14. ,„ ^^Â _^^^^ 2A 1 11. + cdt a = 2 cosec 2<(. cos<n-sin''n = cos2«. 13. ; 1/,, 2 = sec 96. 1 . 2rt = ^9. A 1-cosA .— = tan7r^. - —A 2(9. . -tan A. sin 12. standard formula.'] find sin 2A. 1 + cos 2A . iniporlaiU, o ,''-4 1 type are • Prove the following identities ^ XI. d. c.ntmples 'printed in more promineiU [77/6' [CHAP. = „. i„ 7 1 _. ^. ^cos26'. cos 26). v-^v>,— -= sec' 6 COt a - tan a cotJ-tan.l • TO ~r .4 ^ cot + tan ^ ' ,î^ cot2^ 19. 1A ,(9 = „2cos2-,. 2 2-sec2^ ;-sec2^ • 15. +l , cot^ to-j A — „, 21. „„ 23. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight k' — — sec(9-l sec .- = 2 COt 2a. =^ ^ ^^ • • = sec 2.1 ,, , 1 „^ = „2sm. ( —— cosec2^-2 cosec' „^ .,-i,~-=cos20. »a cosec' 132/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight FUNCTIONS OP 3 J. XI.] Pnive the following identities : 24. fsin-g +cos^y = l + sin 25. (sin ~ 2 cos 2a 26. 1+sin 2a COS 2a 27. - sin 1 2a ) = tan (45° = cot (45° -a). 8^ = 8 sin A A. ^^~ ^^^ '^^^ 2 105 — a). A 2A 28. sin 29. cos4J=8cosM-8cos2^ + l. 30. sin 31. cos2 32. tan (45° (45°- ^) = 2 tan 2A. 33. tan (45° 125. ^=1 -2 sin2 a) (j - cos Ub cos - ~\ cos 4A. . (^- a\ = sin sin2 + ^) -tan (45° + J) + tan -J) = 2 2a. sec 2.4. Functions of 3A. •sin 3.4 =sin(2/l + ^)=sin2.rlco.s.4+cos2.4 =2 cos2 ,4 sin A + (1 sin - 2 sin^ J) sin A = 2sin.4(l-sin2.4) + (l-2sin2.4)sin = 3 sin A - 4sin^J. Similai-ly it may tan 3.4 =tan = 4 cos^ ^ (2.4 ^ we obtain tan 2.4 = 3 cos A + .4)=,^^ ^^ +^Â-^ '^ 1 by putting .4; be proved that cos 3.4 Again, A 2tan^ ; ' - tan 2A tan A l-tan2.4' on roduction tan 3.4 = 3 tan I .4 -3 - tan^ tan2 A .4 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 133/467 . 8/12/2019 . . . Elementary Trigonometry by Hall and Knight KLEMENTARY TRIGONOMETRY. 106 [lHAP. These formulaj are perfectly general and may be applied to cases of any two angles, one of which is three times the other thus cos 6a = 4 cos^ 2a = 3 sin sin 9.1 To pad 126. the value . . • . • . .-. A Since 18° is = 9(r ) 3.L ; — A (which A = 4 cos^ .1 3 cos A not equal to zero) cos is A = 4 cos^ ^-3 = 4(1- sin^ .4) - 3 2 sin ; 4sinM+2sin.4-l = 0; sm.4 = , . .-. 2.-l . 2 sin Divide by cos so that = sin (90° - 3.1 = cos 3 A sin 2vl • . ; 18°. of sin = 90°, Let A =18 , then 5.4 — 3 cos 2a 3 A - 4 sin^ 3 A -1+^5 ^— -— \/4 + 16 ---g -2 + an acute angle, we take the positive sign v/5 - — sml8° = ^^— o .-. : 4 Ksample. cos Since Now Find cos 18^ and sin 18 = r)4° >^/l and co.3r,o - sm2l8° = A / ;J6° V 54°. 1 —y— = ^ ^.y Ih . . 4 are complementary, sin 54° = cos 36 . = l-2sinn8-l-î^-^) = >Â+l. 4 16 .-. sui 54°—--— — . 4 EXAMPLES. XI. e. cosJ=.^, findcos3J. 1. ]f 2. Find 3. (fiveii sill 3.1 tan .1 wlieii sin .-3, find A — tan 3 - . 3.4. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 134/467 . 8/12/2019 . Elementary Trigonometry by Hall and Knight FUXCTIOXS OF 2A AND 3 J XI.] Prove the following identities SA sin , cos SA 4. -. - 3cosa + cos3u sin -, A cos 0-. 3 sin a 6. cos^ a jj ^ _ A ^—- = cot''a. — Sin — cos The *127. —A cot -:r-, - ^4 sin3a + sin3a i ] , --=cotci. -, cos^'o sm — cos3a a -^ = siu54°. + sinn8° = cos-36° 11. ^-3 + sin 3a sin^ a sin 18''+sin30° cot^ , S cot- 7. 3a 3a , „ ,., „ cot3.4=—- 5. cos a 9. : =- - 107 . ? 4 10. os 12. 4 sin 18 36° - sin 18° cos .36° == .^ 7^ following examples further ilhisti'ate the formula? proved in this chapter. Example The 1 Shew that . cos^ a + = (cosâ + sin- a)- -— 1 - 3 - - sin^ 2a. —1 - 2. —= .cos^-sin^ iu that right side = . - ; ^ cos A The 3 cos- a sin- sm- 2a. . 4 JOT.Prove -cos'-'a sin'-'a) (4 cos- a sin- a) 4 n 3 =1 = (coaâ + sin2a){cosâ + siu'*a first side E.vample sin* a sec 2.4 - tan 2.-I + sm A 24 1-— = sin 1 ^ cos 2A cos 2 A sin 2/1 cos jr-— 24 , and since cos 24= cos- 4 - sin- 4 = (cos 4 + sin 4) (cos 4 - sin .-J), this suggests that we should multiply the numerator and denominator of the left side b_y the „ , first cos ., side A- = sin (cos ~ 4 - thus ; — sin 4) (cos : (cos • A 4 + sm 4 cos- = 1 - ; -f^ ) (cos cos-4 +sin24 sin cos 2A 24 — 4 -sin 4) ~ -. 4 - sm 4 -2 cos4 sin4 4 -sin-4 = sec 24 .^ , .... - tan 2A. , http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 135/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETllY. 108 Example 3. Shew that [CHAP. ^^-r tan SA http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 136/467 8/12/2019 Elementary Trigonometry by Hall and Knight FUNCTIONS OF XI.] Prove the following identities , „ cot a - 1 1 — 2.1 AND 3A. 109 : sin 2a http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 137/467 8/12/2019 Elementary Trigonometry by Hall and Knight CHAPTER XII. THANSKURMATION UF PRODUCTS AN J) SUMS. Transformation of products into sums or differences. 128. Iii the last chapter B cos AsmB = sin (J. + B), sin A cos B — cos A sin B = sin {A — B). sin and By A cos addition, % l>y we have proved that fih\ A cofiB = ii\n(A + B) + s,m{A- (1); 11) subtraction ^coÂ&\i\B=ii\n{A-\-B)-s,m{A-B) (2). These formulge enable us to express the product of a sine and cosine as the sum or difference of two sines. Again, and By cos A cos £ — sin ^ sin B = cos {A + B), cos A cos B addition, 2 cos sin ,4 sin .6 = cos (A — B). A cos B = cos (A A sin 5 = cos (.4-^) -cos (A + B) + cos (A ~ D) (li) by subtraction, 2 sin + B) (4). These formula? enable us to express (i) (ii) the jjroduct of two cosines as the the product of two sine.s as sum the of two losines; difference of two In each of the four formulae of the previous article it should be noticed that on the left side we have any two angles .4 and /J, and on the right side tlie sum and difference of these 129. angles. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 138/467 8/12/2019 Elementary Trigonometry by Hall and Knight TRANSFORMATION OF PRODUCTS INTO SUMS. Ill For practical purposes the following verbal statements of the results are more useful. B = sin (sum) + sin {difference) 2 cos A sin B = sin {sum) — sin {difference) 2 cos A cos B = cos {sum) + cos {difference) 2 sin A sin B = cos {diference) — cos {sum). 2 sin A cos ; N.B. the last of these formula, Jn difference jtn'i-edes fJif the sum. Kxomple. Example Example ] . '2. ^. 2 sin lA cos \A 2 cos 8# sin cos = sin (differenvf) = sin 9^ -sin (-3^) = sin 9^ — cos 5A = SA -g 1 .j ^ =^ 4. + sin = sin 114 + sin 3/1. 6^ = sin (36 + 66) - sin — F.xa mple (s«»t) 2 sin 7o° sin 15° + (3^ - fii?) sin 30. oAX) /3A -_ + cos — - — + 5A\ fSA i jcos ( V- ( j {cos 44 + cos (- A)} (cos 44 + COS A). = cos (75° - 15°) = cos 60° -cos 90° - cos (75° f 1 5°) 1 After a little practice the student will be able to omit some of the steps and find the equivalent very rapidly. 130. l Example 1. Examyile 2. 2 cos I sin (a - -7 + 2/3) ^ ) cos cos (a — -d\ = + 2/8) ==^ http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight { cos sin 2a + + cos 20 sin (sin 2a - sin ( = cos 2^. - 4/i) ', Aft). 139/467 . . 8/12/2019 . . Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 112 EXAMPLES. XII. [CHAP. a. Express in the form of a sum or difference *^ ^ 2 sin 3^ cos 1. u 3. 2 cos iy 5. 2 cos 5^ sin 40. 2 sin 34 u 6. 2 sin 4(9 8. 2 cos 9^ sin sin 24. cos 86. 2 sin <^ 2 cos 2a cos 1 In. 10. 2 11. sin 4a cos 7a. 12. sin 3a sin 19 13. cos — *^ 14. Z^ '^ 9. ,_ . \/ 9(9 4 — sin 3<9. 20 _ 2 cos 2/3 50 — cos (a - V ' 2 sin (2^ + ^) cos {0 20. 2 cos 21. cos(60° + <^) 5a sin 10a. a. sin (^f 16. 30 sm - sm —- 18. 2 sin 3a sin (a +/3). .. I ^). 19. (3(9 .sin 7(9. sm 54 cos — ^ ... — 74- 34 . sin 2 cos -— cos 3 3 15. ^17. ^ cos 54. *^ 4. ^7. f' J 74 2 cos 6^ sin 3^. 2. d. - . . 4 4 - 2<^). — - 'ZCp). 2(^) I + a)sin(60' Transformation of sums or differences into products. 131. Since and 1)V .sin (4 .sin (4 addition, .sin(4 1 )y cos ^ + cos 4 sin .5, cos B - cos 4 sin + /J) + ,sin(4-75) = 2sin4 cosJS + i?) -sin (4-^) = 2 sin ^ ; (1); .subtraction, sin (4 Again, and By by +5) = sin 4 — B) = sin 4 = cos 4 cos (A — B) = cos 4 cos (4 -f -5) addition, cos (4+i?) + cos cos cos 4 ^ (2). 5 — sin 4 sin 5, cos .B + sin (4 -7i) 4 sin ^. = 2cos4 cos i? (3); sul)traction, cos (4 + B) - cos (4 - .B) = - =2 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 2 sin sin 4 4 km B sin(-7i) (4). 140/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight TRANSFORMATION OF BUMS INTO PRODUCTS. XH.] A+£=t', and A-B^JJ; Lot then By -1 = — ^, and sin 6 + sm — sm C cos C sni I) x/ =— ^^^ B in =2 sui —— =2 cos —-— J and siiWtituting for 7i — the furnuila- wc obtain - we —-— sin cos - , , , . In practice, it is more convenient to quote have just obtained verbally as follows tlie formu- : of two sines = 2 sin {half-sum) cos {half-difference) ; difterence of two sines = 2 cos {half-sum) sin {half-difference) ; two cosines = 2 cos {half-sum) cos {Jialf-difference) ; sum sum of difterence of two cosines = 2 sin {half-sum) sin {half-diference Exiunpli: I. sin llfJ + siu 6<» Example 2. sin 9.4 - — = 2 sm ^2 sm lA = 2 Example Example o. 1. cos sin lO^^cos'itf. <Â cos 9-^ cos -^ ^2 cos -jj- , + 7A ^ A + cos Â = 2 . revei'sed) -- cos ^ = 2 cos 8.4 9^ . sm A — 7.1 'A .^ sin A. ''-^ cos cos / I - -^r lA -^ cos 7U^ - cos 10^:= 2 sin 40° sin ( . - 30 ) = -2siu40°sin30'_ I (4), . . la? — (:i}, -- cos - iî), (^2), — — COS —-— B-G G+D „ „ sm — -— sin —-— - cos B= 2 + cos i> = 2 cos 132. IIH http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight sin 10= 141/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 114 EXAMPLES. Express in tlie XII. [(JHAP. b. form of a product 1. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 142/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight TIIANSKOKMATION XII.l Tho 134. rUODUCTS AND Ob' 110 SUM,s. bo studied with great following exaiiiplcs sliould care. Example Prove that 1. sin 5,4 The first side + sin 2A - = (sin 5^ =2 sin A = sin - sin ^ + sin ) cos 3^4 sin 2 A first side = ^ (cos =- l). — cos 3^ cos 20. - sin Ad sin 3^* (cos 6 = cos3^ Example 2^1 Prove that 2. cos 2^ cos The A (2 cos 6 A +1). + sin 2 = sin2^(2cos3.4 + Example '2 cos i- M - cos o0) - ^ (cos 0) + cos 50) cos 2^. Find the value of 3. cos 20° The expression = cos 20° + cos + = cos 20° + 100° + cos 140°. 100° + cos 140°) (cos 2 cos 120° cos 20° cos 20° =3cos20°+2( --j = cos 20° -cos 20° = 0. Example sin (/3 i. +7 - Express as the product of three sines a) + sin (7 + a - /3) + sin (a + /3 - 7) - si n The expression =<2 sin (j3-a) 7 cos = 2 sin 7 {cos =2 sin (2 sin —4 sin a sin 7 + 2cos (a+^) {§-a)- (8 j3 cos (a sin (a ( + (i + 7). -7) + ^) sin a) sin 7. cosy as the sum of four cosines. The expression = 2 cos a { cos (/3 + 7) + cos (j8 - 7) Example 5. Express 4 cos a cos = 2 cos -cos (a _ cos (a h /i + 7) + a cos cos (a - + ji + 7) + cos (/3 -1- /i (/? + 7) + 2 - 7) 7- /3 a) + cos -f- cos o cos (a + /3 cos (7 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight i- (fi - 7) a /:<) ~ 7) -) cos (a - H- cos (a ji + (i 1 7) - 7) 143/467 . . 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 116 Example Fimt Prove that 6. siu'-5.r - sin 3.T = + sin (sin 5.r [CHAP. 8x sin sin 2x. sohition. sin'-'5.T - sin 3.r = (sin = (2 = (2 — 5.t 3a;) sin 4a; cos x) (2 cos sin 4.C cos sin 8.C sin 4.r) (2 - sin 4.c sin 3.r) sin x) x cos x) 2a-. Second nolutton. sin sin 2.r 8a- =- (cos 6a- =- {1 - 2 sin^Sx - (1 - 2 sin^Sx)} = - cos 10a;) sin^Sx - sin^Sa;. Third solution. By using the formula of Art. 114 we have at once sin 5.i- - sin- Sx — sin (5.T + 3x) sin (5x EXAMPLES. Prove the following identities ^ W ^ + sin 2A - sin 1. cos 3.1 2. sin 3(9 3. cos (9 4. sina-sin 2a + sin 3a =^4 5. sin 3«-i-sHi /a 6. sin .1 -sin + cos +2 (9 -sin 2^ + cos 4.4 cos _ 9. 10. = cos 3.1 (1-2 5.1 - cos a cos sni sin A). =4 + sin 5a — sin a = tan 2a + cos 5a + cos a , — 3a — COS — 7rt siu .la cos sin 3.1 cos- .1. . 2a. 4a+sin 5a , a+sin 2a + = tai cos a + cos 2a + cos 4a + cos 5a sin c. = sin 3(9(1 -2 cos 2^). 5(9 = cos 2(9 (1 + 2 cos 3(9). + sin sin 8. 1 Su. „ ... cos7^+cos3^-co85^-cos^ v = COt2». i sm 70 - Sin 36 - sin 5^ + sm . v, ; cos 3,1 sin 2/1 sin 2x. 5(9 sin 2a 7. XII. 8,i- : + siii iOo=4 sin 3.1 - 3x) =^ sin . - cos : ;; 4.1 sin .1 — cos http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight -J, 2.4 sin A. 144/467 . 8/12/2019 , . Elementary Trigonometry by Hall and Knight TRANSFORMATION OF PRODUCTS AND SUMS. XII.] Prove the following identities : 12. 3A= — sin 2 A sin sin 4(9 cos 6 - sin 36 cos 20 = sin 6 cos 20. 13. cos 5° 11. cos 5^4 cos 2 A — — sin 25° 117 cos 4 A cos ^1. = sin 35°. [C7sesin25°^cos65°.] + cos 65° = s_f2 cos 20°. 80° + cos 40° -cos 20° = 0. 14. sin 65° 15. cos 16. sin78°-sinl8° + cosl32°=0. 17. sin^ 5 18. cos 2 A cos 5^4 19. sin (a + i3 + y) + sin(a-^-7)+sin(a + ^-y) + sin (a — yS + y) = 4 sin a cos ^ cos y. 20. cos(/3 + y-a)-cos(y + a-/3)+COs(a + ^-y) — cos (a + j3 + y)=4sin acoa ^ 21. ^ 22. A — sin^ 2 A = sin 7^ = cos^ — sin 3^. sin^ -— sin2a+sin2;3 + sin 2y-sin2(a+^+y) ^^ =4sin(y8 + y)sin(y4-o) sin sin y. (a + ^). cosa + cosfJ + cosy + cos(a + ^ + y) =4 , cos +y —^' /3 cos y ^—+ a a+/3 Jt Jt Ji J +A -A) = sin 23. 4 sin 24. 4 cos 6 cos 25. cos^ + cos('y-^Vcos('y 26. cos2 sin (60° ) (^ + e\ sin (60° cos — cos . 3 A. (^-e\= cos 36. + ^')=0. A + cos2 (60° + .4) + cos2 (60° - ^) = | [Pif ( 2 cos- ^ • = 1 + cos 2 J . ] 3 + sin2(120° + .4) + sin2(120''-^) = -. 27. sin2^ 28. cos 20° cos 40° cos 80° =- 29. sin 20° sin 40° sin 80° = o ~ aJ3. o H. K. E. T. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 9 145/467 , 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARr TRIGONOMETRY. 118 [cHAP. Many interesting identities can be established connect135. ing the functions of the three angles A, B, C, which satisfy the relation A + B + C=180°. In proving these will it be necessary keep clearly in view the j^roperties of complementary and supplementary angles. [Arts. 39 and 96.] to the given relation, the sum of any two of the angles su^jplement of the third; thus the From sm{B + (T) = sin A cos {A+B)= (C+ A) = - cos Z? cot A = - cot (B + 0). tan sin C= sin (^ + Z?), P A Again, ment -^ of the tan B, + -^ + sum f -^ = 90 that each half angle , so of the other two , tan 2 B+C ^-- :— liA+B+ C^ 180°, sin 2 A The first side ^2 sin (4 = 2. sin G cos (A - B) —2 sin C {cos (^4 , prove that + sin 2B + sin 2 C = 4 + B) cos .A - cot ,0+A />' f 1. the comple- is thus ; C A+B -sm^. Example C, = - cos (C+A), . cos - cos is sin A B sin - B) + 2 sin (.4 +2 sin C -B) + cos C cos sin (.' ( cos '. C C' j ==^2amC {cos{A-B)-Qos(A+B)'i —2 sin CX — i sin A Example If 2. +B is sin B sin C. prove that the sujiplement of C, 1 - tan />' tan C. we have (A+B)- -tanC; — A+ B = tan A tan B tan .: l)y sin + tan B + tan C = tan A tan whence B A A+B+ C = 180°, tan A Since A 2 sin tan r;. ., - tan t . ; multiplying up and rearranging, tan A -I- tan B + tan C — tan A http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight tan B tan C. 146/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight Sll] Example A+B+ If 3. cos The WHEN A+B+C=180\ RELATIONS C^ 180^ prove that ^ + cos iJ + cos C=l + 4 sin— sin— -2 cos —^— cos —^ =2 — cos first side ., =1+ . =1+ A-B - C . sin ^ -- hcos ^ , „ 2 sin . C — / (cos A-B , A — . sin B . . sin - siu 1 '= ISO', in-ove „C sin 2 A - siu 2B + sin 20= 4 2. sin 2 A - sill 2i/ - sin A+B —^ — C . XII. d. that 1. 3. C cos sln EXAMPLES. U'A + £ + . , +1-^2 sm - = 1 + 2 sin— I2sin— , . — sin sin- ^^cos-^- - sin-j 2 —i+4 119 20= cos J --i sin sin J B cos cos C'. B cos t' sinA + .smB + sinC = 4cos4^ cos?cos^. z z 4. sill .1 5. cos + sTii 7. — ' sin i , s;iu — ccs - i B O J - COS B + CCS C= 4 cos ~ sin — cos ^ sinif „• B - sin O ~ 4 si sin . - L v( C-sin B~. +— j5 ^l+sm —Cri = tan —A'2 ta n —O2 +—sni tan [Use tan i~; ^ 2 tan ^ ' - ^ + tan 2 — .;— — cot ,^ , tan rt«rf 2+ tan ^ tan 1 = thcrcfurc tan ' ' tan 1. V -I.J 9-2 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 147/467 . 8/12/2019 . Elementary Trigonometry by Hall and Knight ELEMENTARY TKIGONOMETKY. 120 If A -\-B+C-^ 180°, prove that B + cos A - cos B + cos C ^^=taii 1+ cos .4 + cos i?- cos C 2 1 ^ , 9. cos 10. 11. 12. ; cot -„ x. C ^ . 8. [CHAl*. -- 2 B cos C + 1 = 0. cot BcotC+ cot C cot A + cot .1 cot ,5 = 1 (cot B + cot 6^ (cot 6^+ cot .4) (cot A + cot B) = cosec A cosec Z» cosec C. cos^ ^4 + cos^ B + cos- C+ 2 cos J. cos B cos 6'= 1. 2A + cos 2B + cos 2C+ 4 cos A -+ sin^ 14. cos2 2 sin2 - + sm^ -= 1 - 2 sm - sin A + cos^ 2B + cos^ 2C= 1 + 2 cos cot^ + cotC cotC+cot^ tan ^ + tan tan C 16. (sin J. *136. The +sin jB+sin g »' 2.4 cos g • 2B cos 2C. cot.4+coti?_ C+ tan A tanJ. — tan tan^ + tani^ + tanO' = l + cos24.] 2cos2/( [C7se 13. cos C)'-^~ 2 cos — tan — tan 2 A + tani> 2 cos 2 B cos C following examples further illustrate the IbrmulcC proved in this and the preceding chapter. Example „,, The „ 1. , first ., side Prove that cot cos (4 =-^ + )-, 15°) + 15°) - tan sin ^-^^ 15°) (A - 15°) 4 cos = . 2^4 oj + f • — (^-15°) ) ^Vm . + — {A+ sin (A cos _ ~ (.4 cos (A 15°) cos (A sin (^ - - 15°) 15°) + - sin (A 15°) cos (/I + 15°) sin (.4 - 15°) -15°) + 15°) + (^-15°)} sin (^ + 15°) cos (^-15°) cos {(^ _ ^ 2 sin 2 cos {A 4 cos + it is tlic 15°) cos (^ 15°) 2 cos sin 2.4 2A + sin 30° * l In dealing with expressions which involve numerical angles ustially advisable to effect known - _ ~ 2A ~2sin22t + Note. 2A some simplification before substituting values of the functions of the angles, especially if these contain surds. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 148/467 8/12/2019 Elementary Trigonometry by Hall and Knight MISCELLANEOUS IDENTITIES. XTI.] Example cos li 2. A+B + C = B A prove that ir, G — ir-A — + cos — + COS 17 = 4 COS 2 2 The second , = 2 cos = ( I V-A W-A — p —7- —+ A IT TT cos 4 -+ — A + — cos -— „ A\ ^ — +2 TT . cos COS I + i3) sin (a 4 cos COS B+ — cos C — -C —4J— . — - —— — — cos B-C TT-A „ 2 cos h ir COS l-cos^ - - 4 4 — cos B-C —j- B —+- C j Prove the following identities cos (a -. 4 *EXAMPLES. 1. TT-B ~ cos ir cos COS 4 2 = 2 COS —j— side 121 . XII. e. : - /3) + cos (/3 y) sin (/3 - y) + cos(y + S)sin(y-S) + cos(S + a)sin(8-a)=0. 2 sin(3-y) sin(y-a) ^ sin/Ssiny sin(a-j3) ^ sin ' y sin a sin sin a sina+sin^ + sin(a+g^^^â sina+sin/3-sm(a+i3) 2 ^Q ^ ^,g_ 2 4. sin a cos (/3 + y) -sin ^ cos (a+y) = cos y .sin (a -^). 5. cos a cos (/3 + y) - cos )3 cos (a + y) = sin y sin (a - 6. fcos^ —sin 7. If tîn 6 =- a .4) , (cos 2.4 prove /3). — sin 2J) = cos.4 -sinSJ. tliat a cos -26 +h sin 2^ = a. {See Art. 124.] 8. Prove that sin 2 J +cos 2 A (2+tan^)2-2tanM l 9. 10. Prove If J + + tan2.4 4tan^(l-tan2^) tliat /? sin 4/1 (l+tan2 J)2 = 45°, prove that (l + tan.4)(l+tan5) = 2. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 149/467 . 8/12/2019 . Elementary Trigonometry by Hall and Knight flfLRMENTARY TRIGONOMETRY. 122 Prove the following identities 11. eot(\5'' 13. : - A) + tan(}5'' + A) = -^^^:^1 - 2 sm 2 J rat(1.5° + J) + tan(15° + J) = + 30°) tan (A A+ 14. (2 cos 15. tan (^~y) tan I) (2 cos + tan - 30°) (.-1 A- (y-a) cos - 1 +2 - sin y) + sin cos^ (jS 18. cos^ o 19. sin^ a 20. cos If 21. Cos A - +COS A COS 2 „^ 23. -- sin 2a tan 1 - y) tan (y - «) tan (a - /3). - ^) . sni /3-y —— y-a . ^ sni -^t— a-j3 . sm -- - _ = 0. a) 2 B — 6' - j3). r: C + COS — = 4 COS /3 -. - C -- ^ ^ 2/3 :~r + sin 2^ tan y -, , n + tan cos 7T-B 4 , 1 +4 ^ —+ C ir -- A .n-A sin COS ^ . sin 4 . A n-B sm —47 . tt-O -— . 4 that + sin - , +A , 2 . + sin C+A A+B cos — cos — 4 4 4 B —+ 2 + y — '-, shew —. (/3 (a +COS- =4cos . .sm 2a 25. B .A---+SIII B +S1II 1.7 -. 24. = 2 cos 4/1 + , shew that 2 sill Tf (j+/j 1) + sin2 /3 + 2 sin a sin /3 cos (a + /3) = sin^ (a -f /3\ 12° + oos 60° + cos 84° =- cos 24° + cos 48°. 2 22. 2^ + cos^ (a - /3) = 1+2 cos O - y) cos (y - a) cos (a cosacos(3cos(a + ^) = sin2(a+^). + cos^ /3 - 2 A+B + CÎSO cos cos + tan (a - j8) (y-n) + sin - y) + cos'-^ (y - 2.4 2 cos 2 A 24 - 1) (2 cos +4 17. . 24 + ^3 sin 1 -=• =tan 16. XII. ' ' 12. [OHAP. '. 2y ^ ^ COt « 2y y tan « sm -| C< it d. tan u tan http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight /:< 1 150/467 8/12/2019 Elementary Trigonometry by Hall and Knight MISCELLANEOUS IDEXTTTIES. XII.] EXAMPLES. Prove the following oosec {a ^ / r, '^ ^ ^, + ^)= / a\ ^ ' ideutitie.s — —+ cota cosec a cosec ^ r 1 -?. ; + tanatan/J sec n sec - cot . ' cotyS = tan /3 3. losec 4. 2cos2(45°-^) = l+sin2J. 5. 2 cos 2.r cosec 6. cos(^+7?)cos(.l-i?) + 7. 1 2 sin- A ). 8. E.Kpress cot 2.1 in terms of tan .4, -16 + cos 4.4 2(9 6. Zx = cosec x — cosec =2 f. Examples on Chapters A'l and XII.) (East/ Miscellaneous 1. Xn. 122. cos 2.4 l - ( 1 Zx. = cos2J+co.s2^. ^ and t;in 2.4 in terms of cnt.l. A= ^ If tail 3^ 9. \\\ sin prove that .4+tan / 4< .4 =:; , .sec .4 (ii) ; + tan .4 = — {\+tf - v-. sin 3.4 sin 2rl .sin 0=^-28 — sin .4 cos ,3 ==-6, find the value of cos(a + /3). of cosines find a + /3 to the nearest minnte. Thence from the table and Check the result by finding « and /3 separately from the data. 11. 12. 13. tion -• If a=^ + y, .shew that sin(aH-/3 + y) + sin(a + /3-y'i + sin(a — /3 + y) = 4 .sin o cos /3 cos y. If Prove that cos 57^ + sin 27° = cos 3 , and verify the rela- by means of the Tables. 14. Express 4 sin 5a cos 3rt cos 2a a.s the sum of three sines. By means of the Tables find ai)proximately the numerical value of the expre.s.sion 4 cos 2U' cos 30^ cos 40°. 15. [Fii-st express the product as the sum http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight of three cosines.] 151/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTAKY TRIGONOMETKY. 122g Find the smallest value of 6 which 16. sin A6 cos p satisfies = 7 + sm ^ cos — the equation . Prove the identities 17. (i) {x tan a +3/ cot a) (ii) cos (iii) cos (2a /3 cos a Check cos 2^. cot a +3/ tan a) (.f - /S) = cos'^ a - sin^ (a = {x-\-y)- + 4i.y cot^ 2a = 8, find the results ; - ^) + cos 3a + cos 5a + cos 7a = - sin If cos ^ 18. *i [CHAP. XII. 8a cosec a. the numei'ical values of sin 2^ and by first finding 6 from the table of cosines. If 2.1 20. Prove that -,J— (i) sm ^ ^ +B=m\ 19. 21. ?Z?^=:4; 2 aJ ^^'T^ sin 54° (ii) ( 1 - sin If .4 23. If 1 prove that 5 sin C) = cos'' B + cos^ C. Prove that — cos^ A cos 25 = sin- 5 — cos^ B cos Prove that tan 50° - tan 40° = 2 tan lation by using the table of tangents. 25. 26. = sin 162 + sin 30°. A + B+ C= 0, prove that + 2 sin i? sin C cos ^ + cos- A = cos^ B + cos^ (7. siu^ ^4 f ' + 2? + C= 180°, prove that 1-2 sin B sin C cos ^4 + cos^ .l,=cos2 ^+cos2 C. 22. 24. A= cos 10 10 5 + C= 180°, If prove that cos If cot ^ = -5 find sin 2^ and cos 26. 10°. '2, A. Verify the re- Check the values as in Ex. 18. 27. 6 ine;ins of the Tixbles, find tlie By which satisfy the equation •362 cos ^ [From the two smallest values of + sin (9 = 1. = .362; then (90° - a).] = = sin cos o sin (a + 6) table of tangents find a such that tan a the equation may be written http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 152/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight CHAPTER XIII. RELATIONS BETWEEN THE SIDES AND ANGLES OF A TRIANGLE. In any triangle 137. the opposite angles; that the sides are proportional to the sines is, _ a ABC be Let the triangle AD From A draw the opposite side _ 6 sin^ sin J. (1) of c sin C acute-angled. perpendicular to then ; AD = AB sin ABD = c sin B, AD = AC sin ACD=b sin and b sin .*. that C=c sin b is, C sin Let the triangle (2) Draw AD B, c 5 sin C; ABC have jjerpendicular an obtuse angle B CB to produced; then AD = AC sin A CD = b sin C, A D=-AB sin ABD and = c sin {\m° - B) = c n\n B b sin C^c sin B; .•. , equal to m 1 hus = -; IS, In like is b . that B sin manner -. — sin -. A it may = c —TV sin C -: be proved that either of these ratios . a -. sin -. A = b -. c Tj sin /J = r—^ - sm C http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 153/467 8/12/2019 Elementary Trigonometry by Hall and Knight KLEMENTARY TRIGONOMETRY. 124 To find an expression for one side 138. terms of the other two sides Let (1) C Draw BD then by Euc. and (- . {c) of a triangle in the inclnded angle '{€). be an acute angle. perpendicular to AC; 13, ir. A B^ = BC-^ + CA^-2AC.< . [chap. 7> = a^ + 6- — 2ft a CO.S C = rt^ + — âh cos . ( '. ft - Let (2) (' be an obtuse angle. Draw BD perpendicular to produced; then by Euc. ii. 12, AC AB^ = BC^+CA'^ + 2AC. CD; .-. = a^ + h'^ + 'ih. a C0& BCD c ^ = a2 + + 2a6 cos (180° = a^ + b^-2abG0f^C. C) ft2 Hence in each case, Similarly it c'^ K — a'^-\-lr may be shewn - b 2aZ)cos C that a2=ft2^c2-26ccos.^, and ?>2^^2 From 139. eos , î c^ = h^ +—^ _f. (-^2 _ the formulae of the last — a'^ 2ftc ; cos „ n— c- eos B. 2^0^ article, + a^ — V^ , cos : 2ca ' we obtain ( = —+— a? ft- -. - c^ . 2a6 ' These results enable us to find the cosines of the angles when the numerical values of the sides are given. To express one 140. side of a triangle in terms of the adjacent angles and- the other two sides. (1) Let ABC be an acute-angled triangle. Draw .4i> perpendicular to BC; then BC=BD+CD = ABcos ABD+A that is, cos a = ccos i^-fftcos .1 CD ; 6'. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight q 154/467 ' 8/12/2019 . . Elementary Trigonometry by Hall and Knight RELATIOMS BETWEEN THE SIDES ANP ANGLKS. XIII.] AD Draw produced ; perpendicular ,/ have an obtuse angle BO to A - cos A CD ; = rcos5-Z)cos(180°-6') = c cos B+b cos ( Thus C. then BC=^BD-CD = ABcohABD .-. ABC Let the triangle (2) 12. a^b in each case Similarly it may B '. cos 0+ c cos />. be shewn that ( h = c cos A +a cos and ', câ cos B-Vb cos A Note. The formulae we have proved in this chapter are quite general and may be regarded as the fundamental relations subsisting between the sides and angles of a triangle. The modified forms which they assume in the case of right-angled triangles have already been considered in Chap. V. it will therefore be unnecessary in the present chapter to make any direct reference to right-angled triangles. ; The and 140 have been established independently of one another they are however not independent, for from any one set the other two may be derived by the help of the relation J.+.B + C=180°. *141. sets of formula? in Arts-. 137, 138, ; For instance, suppose we have a sin then sinc^ sin . . b A 1 == B — sin -. ; 1 =^ a • Siinilarly, . B sin a —b C — sm A sin ,, cos C+ —. .4 b . e _ C A=sm(B+C^ = sin B cos C+ sin sni .-. sin cos ^ — <^ (7 -f cos 137 that j:)roved as in Art. 7 cos ( cos ' B ; „ o T, « ; a. cos f -f c cos Tl. we may prove that b — GQ.os.A-Va cos 6', ai id e =a cos 5+6 cos Multiplying these last three equations by a, tively and adding, we have .4 ?), -c resi.>ec- = 2a.6cos(7; c2 = a2 + J2_2aftcosC. a2 + t2_c2 .-. Similarly the other relations of Art. 138 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight mny b.> deduced. 155/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 126 [fRAP. Solution of Triangles. When any three parts of a triangle are given, provided that one at least of these is a side, the relations we have proved 142. enable us to find the numerical values of the unknown parts. For from any equation which connects four quantities three of which are known the foiu-th may be found. Thus if c, a, B are we can given, find h from the formula fe2 and if = c^ + a^ — 2ca cos B we B, C, h are given, from the formuLi find c h c sin We may remark that ; sin C B the three angles alone are given, the if formula _ a sin A b sin c B sin C enables us to find the ratios of the sides but not their actual lengths, and thus the triangle cannot be completely solved. In may be an infinite number of equiangular satisfying the data of the question. such a case there triangles all 143. • Case To I. a solve triangle having given the three 'es. The angles cos , ^4 and = b^ + c^ B may — a^ ^r^ 1. C is If be foimd from the formula? , , 2oc then the angle Example A and cos _ B= c^ + a^ — b^ 2ca ; known from the equation C=180° — J — B. a= 7, h — 5, c = 8, find the angles A and J>, having given that cos 38° 11' = -— 14 62 con A =- cos 7?: + c2-aS 2bc 52 + 32-72 ~ 2x5x8 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 40 1 156/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight SOLUTION OF TRIANGLES. XIII.] Example. triangle Find from the Tables the greatest angle 2. whose sides are Let a — 6, ^ 127 = 13, the greatest side f 13, fi, = ll. of the 11. Since the greatest angle the required angle is is opposite to B. And ''' - cos 84° Case II. induded angle. To 144. A be given solve now cos then C is known Example. If obtain _ jO = 13'. a triangle having given two = 6^ 4B (fi — 2bc cos a'^ — b'' . 2ca = 3, b A — 7, „ or sin JJ = , from the equation « sides a7id from either of the formulaj —+— c^ 18'. then a can be found from the formula ; a- We may 95° is - '^^^^ from the Tables; 47', Thus the required angle 6, c, -12 132 £ = 180° -84° 47' = 95° .-. Let + 62- ^-^^^ = -¥>nnr6- = 27inr6= = the 112 c='+a2-6-= ^ 6sin^l a C= 180° — A ; - B. C'=98°13', solve the triangle, with the aid of the Tables. c^ = a^ + h'-^-'2ab cos G = 9 + 49 - 2 X 3 Now X 7 cos 98° 13'. cos 98° 13' = - sin 8° 13' = .*. - -1429, from the Tables, c2=58 + 6 006 = 64, approximately; c .-. cosB= c2 + a2-fc2 2ca 64 L'=: 180° = 8. + 9-49 24 1 ~ 2x8x3 ~2x8x3~2' .-. .-. [Art. 98] JJ = 60°. -60° -98° 13' = 21° 47'. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 157/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 128 145. and a ( ', triangle havhifj <jireii two aiiyles a be given. The angle A is from c a solve side. Let B, and To Hi. C'AiÊ [CHAP. found froja A-^lSO^-B-l'; and the sides b , a —AB sin —— ; J- sni Example. If .1 a sin , and = 105°, C = 60°, c —— ; i- sin i = 4, C . A solve the triangle. I>':=::180°-105°-60°=15°. ° fc _ '^' (7 sin _ sin£ ~ 4 sin 60 sin 15° f .-. 4 sin 105° _ V3 + 1 sin 15°~ 4 sin 75° ~ sin 15° 2^2 rt^ 4 (2 XIII. 2. 1 3. If 4. lf« = 25, /' /^ t( = 5, b fii c- Hnd id , The (3 6. (Solve 7. Solve the triangle the triangle *8. If a = 8, *9. If the sides arc as -^ 5, c 2, 2^, . J?., when a = ^3 + l, when a — ^/2, = ^/l 9, b b with an asterisk.) ' //'f.'' I sides of a triangle are /> ^ (^ = biJZ, = 5, find the angles. 6 = 31, c=7x/2, find.4. 5. ' a. marked {Tables must be used for Examples = 15, = 7, c = 1.3, - 3, = 5, f a = 7, ^3-1 + ^/3). EXAMPLES. If(( 4(V3 + 1) '\/3-l 2^/2 1. V^ = fi^/2 + 2^6. _ b sin J ~ sin jB ~ .-. _ 4^6 2^21 ~ ^3^1 _ ~2^ ~ 4^3 y'^ ^ ^ ' - ' find the greatest angle. — 2, c~,^6. = 2, o=^3 — 1. find G. 4:7: 5, find the greatest angle. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 158/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight SOLUTION OF TRIANGLES. Xlll.] fi^ 10. = ^/3 + 1 /> 11. Given a = 3, *12. Given 5 = 7, *13. If & = 8, *14. If a = 7, A^ t = 2, If « (.- = 6' (.'= , 60\ find 129 -•. = b, B= 120', liud b. c=G, .4 = 75° 31', tind c 11, = 3, J =93° iy=123° <i. 35', iiiul n, 12', tind />. = 2^/6, = (5- B — lb'. 15. Solve the triangle when rt 16. Solve the triangle when J, 17. Given 18. A UB = 60°, 19. If 20. Given .1=45°, 21. If (: = 120°, c 22. If 23. Given {a + b + c) (b+c- a) = Sbc, find .1 Find the angles of the triangle whose sides are 24. J = 15°, =45°, B = 60°, = 30°, = 72°, = 6 2 ;^'3, = 2, f = ;^/5 + l. solve the triangle. B C= 15°, & = V6, solve the triangle. 5 = 105°, c = s/2, solve the triangle. ^8, a 5 = 60°, = 2^/3, = 3, 6 shew that a = 2, find = 3 ^'3, •S 25. b (; : tt = v'3 + l 2. c. 2^3, ^'6. Find the angles of the triangle whose sides are x/3 +1 2^2 ^3 v/3-1 ' 2^2 2 ' 1 1 26. : />. find + J3, c Two sides of a triangle are included angle is 60° : — Vo-\'2 and r^r r- -77; ^'6 ;- + x'2 , and the solve the triangle. AVheu an angle of a triangle obtained through the medium of the sine there may be ambiguity, for the sines of supplementary angles are equal in magnitude and are of the same sign, so that there are two angles less than 180° which have the same sine. When an angle is obtained through the medium of the cosine there is no ambiguity, for there is only one angle less than 180° whose cosine is equal to a given 146. is quantity. Thus if if sin /1=|, then .4=30° or 150 cos A = i, then .4 ; - 60°. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 159/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 130 IfC = 60°, Example. me 1 rom . & equation we nave = 2^3, B = --^— 7J = SJ2, , —r—2 = ^2' . ; = 45° or 135°. The value i' — 135° is inadmissible, and C would be greater than 180°. To Case IV. 147. solve is tf, If 6, A be given a Z> =- a sides a>id sin be found from the equation ^4. ~>L 5> so that sin a no solution. then = 1, so that < and two values sin^ = I, l, which and B 90^. , so that one angle B= is . 1, for B may These values are supplemcntai-y, acute, the other obtuse. If rt<6, then (1) is is to sin ^1 then be found from sin t)r A, then Thus there impossible. (ii) of 75°. a triangle having given two sin ij (i) sum opposite to one of them. an angle Let for in this case the ^==180° -60° -45° = Thus tind ^. c •ds.1'2 .-. B B— sin sin c [c'HAP. A<B, and therefore B may obtuse, so that both values are admissible. either be acute This is known as the ambiguous case. (2) then A>B; in either cannot be obtuse, and therefore only the smaller value of If a = 6, case B B admissible. is When B Finally, c is may then A=B; found, C is and if a > 6, determined from C=180° be found from the equation c=^ sin -A-B. r-. A From the foregoing investigation it appears that the onli/ case an arnhiguous solution can arise is wJien the smaller of the two given sides is opposite to the given angle. in which http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 160/467 8/12/2019 Elementary Trigonometry by Hall and Knight THE AMBIGUOUS CASE. To discuss the Ambiguous Case geometrically. 148. Let a, towards X CD h, ; AX A be the given parts. Take a line unlimited make z XJ (7 equal to A, and ^4(7 equal to h. Draw perpendicular to If (i) can AX, then CD=b C and With centre not 131 sin A. radius equal to a descril)e a a<5sin.4, the circle. circle will meet AX; thus be constructed with no triangle the given parts. (ii) If AX a=h ain A, the circle will D thus there is a right-angled triangle with the given touch at ; parts. (iii) will cut (1) If a > 6 sin AX in A, the two points B^, These points circle B.y. both will be on the same side of A, when a<b, in which case there are two solutions, a namely the triangles ab[c, AB,C. Ambiguous This is the {•£) The points > h. 5,, In this case there is solution, for the angle Case. B., will be on opposite sides of A when only one CAB^ is the supplement of the given angle, and thus the triangle AB.,C does not satisfy the data. (3) If a = b, the point B., coincides with A, so that there is only one solution. H. K. K. T. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight ~~~- - '' 10 161/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 132 Example. We Given B-45°, c = ^12, sin (7:= .-. C' = J8, 2^3 csinJ5 have h = 60° I [chap. solve the triangle. _JS or 120°, and since b < c, both these values are admissible. which satisfy the data are shewn in the figure. Denote the by A^, A.^ BC^, sides BC.-, by The two triangles and the angles BAC^, BAG,, a^, a,, respectively. A (i) Zî = 75°; IntheA^BCj, 2^2 ftsinî , ^^ ''^ ^3 + 1 '^^^-î^=^--V2=^'(^''+^)V2 (ii) 2^2=1^°; In the A^JJC,, ^2 2^2 = _^- =-^ ft hence .., sin ^3-1 2^-=^/2 U/3 . . , ,> 1). V2 /C = 60°, Thus the complete solution ) ^ _ ygo ^j, jgo . ' ' is ) (^rt=^r) = Given 2. Given 6 = 3^/2, r( Tf C=GO'', 1 , + ^2, EXAMPLES. = V^, ^ = ^ °j 1. 3.* or 120°; ?. ^ a = 2, = 2^3, c = V6, or ^r.-^2. XIII. b. «<^lve tlie triangle. ^'=45°, solve the triangle. Holve the triangle. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 162/467 - 8/12/2019 Elementary Trigonometry by Hall and Knight KKLATTONR BKTWEEN THE STDKS AND ANGLES. XTIT.] If 5. If -5=30°, b 6. If If a .4 = 2, =30°, a 4. c = 5, '133; solve the triangle. = j6, c = 2V3, solve the triangle. 60°, 6 = 3 ^2, c = 3 + V3, solve the triangle. 5== 3 + ^^3, c = 3 - V3, C= 15°, solve the triangle. If ^ = 18°, a = 4, 6 = 4 + V80, solve the triangle. If 5 = 135°, = 3^2, 6 = 2^/3, solve the triangle. 7. 8. 9. 149. triangle Many relations connecting may be proved by means ^ the sides and angles of a of the fonmilse we have established. Example Prove 1. sin then .: {b a - a j^_ Let c) cos =k A ^=k = 2k B- cos -, A cos ^ sin A ~= a sin _ c B sin b=kainB, sin C) cos B+C —-— sin ^ A = 2k sin — . c) _ sin A, (sin ~ {b tliat cos . c B—— C ^ . ' C =k sin C A — B-C —- — cos A— — sm B-C —^ . = « sm^ sm -^ = a sin B-C . Example 2. If « cos- the triangle are in . C A — +ccos- - = C A 2acos2-- + 2ccos*— .-. a :. (1 a + cos +c a .-. b, c C') -\-{a :. sides «, , shew that the sides of a. p. Since Thus the Sb are in + cos ( (1 = 3/>, + cos ^) = Hb, C+c cos ^) = '6b, + c + 6=3fc, a + r=2ft. a. p. 10—2 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 163/467 8/12/2019 Elementary Trigonometry by Hall and Knight KLEMRNTARY TRTftONOMETRY. 134 K.r ample Prove that 3. (//- - C') the _ h — sin yJ A = first = A + cot -. Let nr. . siu (£ + A + — sm^ But the ^t sin [B + C) \ ) = sin A and , cos i H- 9,inA [Art, 114]. . ) ^= - cos [B + C) ; side first ^ C — Q). cot ] ^^, + ' ^ -fc2{sin(/^-r.')cos(B + = - h-) then + 7 C) sin (B - G) ' ; v; COS n^, ( .•. —B = —sin—C^ sin si (sin^^-siu^C'l ^ //•: B-\'{a-~ a^) cot h (^ r- - [c- -; . — side \, -.n A-- [f HAP. sin (sin 2ft I '2C') + C') + -|- • (sin '2C - sin 2,4) + sin 25)} (sin 2.4 ^0. EXAMPLES. Prove the following identities B - sin XIII. c. : 3. + b (sin C - sin A) + c (sin .4 - sin ^ i = 0. 2 6c cos A + ca cos B + ab cos C) = a- + 6- + c-. a (6 cos 6' -c cos 5) = 62-^2. 4. ('t-(-c)cos 5. ^ 1. 2. a (sin ( 8. 9. cos 6 +r a+ ^. 10. ^^- B ?, (o J + (c + a) a siu^ y ( cos 6. C) h = t;os i? + i'a + 6)cos + c sin- ^ = c + a - J 1o - c cos A cos „ 7. I =rtcos )sin .,C •^ '-, . « sin {B - = C'; ^ + 6 + c. 6. | c-h C'=<:/ « sin C tanJ=T b-a cos Y,. C . . • --^+ cos5 cos,4 • + 6sin {C~ J) + fsin {A-B) = 0. im{A-B)_a^-b^ »m('A+B)~ c^ - B) â^-b'^ c^~a^' bsiniC-A) c sin ' http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight {A 164/467 8/12/2019 Elementary Trigonometry by Hall and Knight THE AMBIGUOUS XIII.] [All articles 135 CASE. and examples marked with mai/ be asterisk ait omitted on the first reading of the subject.^ The ambiguous *150. may case be also by (liscii.s.setl linst finding the third side. As before, let a, A b, be given, tiien b'^ , c'^ — d^ c'^-2bcoHA.c + b'^-'a:^ = 0. .-. By + solving this quadi'atio equation in c=b =b c, we ± \/62 cos^ A + a^- cos ^-1 cos A + \/a2— 62 sin2 obtain b'^ ^. When a<bsm A, the quantity under the radical so that there negative, and the values of c are impossible (i) is is ; no solution. (ii) When a = 6sin =b ^-1, the quantity under the radical A< and c fore 5. Hence the triangle ^< acute, in which case A angle cos A. Since sin 1, it a<b, and follows that is impossible c is positive is is zero, there- unless the given and there is one solution. (iii) (1) When a > b sin A, there are three cases to consider. impossible unless A is Also \/a^—b^sin^ that is that acute. A is V«^ — b^ hence both values of two solutions. (2) A < B, Suppose a <b, then c and V^^ - b^ sin^ < Ab, then is In this case b cos real sin^ and as before the triangle \/«^ — ^f b^ ^1 — A is is positive. b^ s'm^ A ; ; positive, so that there are b^ sin^ A>b cos A A > \fb'^ - b^ sin^ A ; ; hence one value of c is positive and one value is negative, whether A is acute or obtuse, and in each case there is only one solution. (3) Suppose a = b, then \/«^ — .*. hence there is c=26cos only one solution obtuse the triangle is H^ sin^ -4 or when A A = b cos ^1 ; is acute, and wlieu .1 is impossible. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 165/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 136 Example. If b, B c, - a^f + («i where Wj, From ii.^ are given, aud + fla) {« [cHAP. b<c, shew that if tanî? = 4b-, are the two values of the third side. « we have But the roots —^ cosi)= the formula - 2c cos i> . a , + c- - of this equation are a^ //- and = 0. hence by the theory of a„; quadratic equations flj (flj .: .: («i - + a.^) {a^ + a^ — 2c cos B and - aj)^ aia2 — c^- b = (« + flo)^ - 4aia2 — 4c- cos- iJ - 4 (c- - + a„)- tan-S = 4c- cos-B - 4 . b'). (c- - b'-) ^ 4c2 (cos-B + sin*£) - + ic^ cos^B tan-i? 4c* + 46^ = 4c--4c2 + 4b= 4fe2. *EXAMPLES. XIII. d. In a triangle in which each base angle solve the triangle. third angle the base is 2 1. is double of the : If 2. is 3, 5 = 45°, (7=75°, and the perpendicular from A on BG solve the triangle. If 3. a = 2, 6 = 4 - 2 ^,'3, c = 3 ^/2 - J6, solve the triangle. - 18°, a = 2, ab = 4, find the other angles. Given 7? = 30°, o = 150, 5 = 50^3, shew that of the two 5. triangles which satisfy the data one will be isosceles and the 4. If ^= 6 other right-angled. Find the third side in the greater of these triangles. Would the solution be ambiguous if the data had been 5 = 30°, c = 150, 6=75? is V^— 7. ami 1 8. ^ = 36°, If 6. rt = 4, and the })erpendicular from (^ upon AB find the other angles. 1, If the angles adjacent to the base of a triangle arc 22^° shew 12i°, If terms of tliat a — 2b and the altitude A = SB, is find half the base. the angles and express c in a. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 166/467 . . 8/12/2019 Elementary Trigonometry by Hall and Knight RELATIONS BETWEEN THE SIDES AND ANGLES. XIII.] 9. Tte of a triangle .sides shew that the greatest angle Shew that COS (6 11. «,sinf-+5J = 12. sni C\ B+— fry { .v^ + 2.v: —-— cosec —-— ] + c)sin^. (6 C - = cos + — 6+c a 2 b , 13. (7 + cos {A - B) cos ^ 1+ cos (.4 -CO cos ~ 14. If c* If (', 1 - + 3.f + 3, 120°. is C+ c (cos B — cos A) = c sin 10. ,o .(•- any triangle in — a) are 2x-4-3, 131 2 (a2 + 62) ffl2 a2 cos A B-C — cos 2 2 ^ 52 + c- + „4 + (^2^2 + 6* = 0, c2 prove that C is 60° or 120°. 15. A /', are given, and if c\^ are the values of the <:, third side in the ambiguous case, prove that if c'j>c^, ,^. (2) h sin A C-,-Co cos^^=^^. (3) c^ 4- c^ / side, By cos, c^ ,\ (4) 16. — C2='ia (1) If -4 . sin 6, — 2ciC.^ cos 2A= 4a^ cos- Cv) +— -^^ = 45°, and ^1 „ 6j — C9 — ^-^— = cos ^1 cos By . . sin =^ be the two values of the ainbigiiuus c^, c, shew that cos 17. If cos .1 that the triangle 18. +2 is cos C : cos ^^,. .1+2 cos 5 = sin 5 : sin (', prove either isosceles or right-angled. r., shew that -, cot .^ cot .^ are If a, h, c are in A. x'4 --, cot 19. BiCB, = — 1 • al.so 111 a. v. Shew that {B-C) sin B + sill C a^ sin i^siu c'^\n {A- sin sin .1 + sin iJ {C-^) (7+ sin A http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight B) ^^^ 167/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 168/467 8/12/2019 Elementary Trigonometry by Hall and Knight CHAPTER XIV. LOGARITHMS. Definition. The logarithm of any number to a given base is the index of the power to which the base must be raised in order to equal the given number. Thus if cv^=N, x is called the logarithm of ^V to the base a. 151. Example 1. Since 3* = 81, the logarithm of 81 to base 3 Example 2. Since 10^ the natural numbers = 10, is 4. 102:^100, lO^^lOOO, are respectively the logarithms of 10, 1, 2, 3,... to base 10. 100, 1000, Example 3. Find the logarithm of -008 to base 2.5. Let X be the required logarithm; then by definition, that (52)^ is, = 5-3, or 5^^ = 5-»; whence, by equating indices, '2x— - 3, and 3^ to base so x= is - 1-5. usually written log„ xY, 152. logarithm of The same that the meaning is expresseda by the two equations From these equations Example. Let log .01 it is Find the value of =X -00001 ; then ( evident that aôga^'=N'. log .qj-OOOOI. -01)^ = -00001 • 100000 VlO^y ?M^--J— .-. When 153. arithms is 2a; ' = 5, and or .r J^= = 102^: 106 2-5. understood that a particular system of login use, the suffix denoting the base is omitted. it is http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 169/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 140 [CHAP. arithmetical calculations in which 10 is the base, instead of logio2, logj^S, visually write log 2, log 3, Thus ill we Logarithms to the base 10 are known as Common Logthis system was first introduced in 1615 by Briggs, arithms ; a contemporary of Napier the inventor of Logarithms. Before discussing the properties of common logarithms shall prove some general propositions which are true for logarithms independently of any particular base. The logarithm of 154. a^ =\ may be. For base For = a; a^ a ; therefore log the base itself therefore log„ a = To find 156. MN be the product Let and suppose let ; a prodtict. a be the base of the system, ay = N. = M, MN= a^ xay = Thus the product whence, by definition, 1. y = \ogaJ^; A-=log„i/, a'' is 1=0, whatever the 1. the logarithm of so that all 1 is 0. for all values of The logarithm of 155. we a'' + i>; logaiI/iV=.r+^ -log„J/+log„#. Similarly, and .so on for Example, log,, = log„ M+ log„ JV+ log„ F any number of log 42 To find 157. 2IJVI' = log (2 factors. x 3 x the logarithm 7) = log 2 + log 3 + log 7. of a fraction. M Let -^ be the fraction, and suppose A- = log„ http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 170/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 171/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 142 [c'HAP. the equation 10-^ = ^y, it is evident that common logarithms will not in general be integral, and that they will From 160. not always be positive. 3154>10^ and <10^; For instance, fraction. •06>10- ^ and <10-i; Again, .-. -2 + a log -06= The Definition. 161. =3+a log 3154 .'. fraction. integral part of a logarithm is called the characteristic, and the fractional part decimal is called the mantissa. 162. when expressed characteristic of the logarithm of any down by inspection, as we The base 10 can be written shew. To determine (i) number greater than It is ttnity. number with two digits a number with and 10^ 10^ integral part lies between and in its integral part three digits ; number with n number to shall now of the logarithvi of an;] clear that a between lies the characteristic as a digits in and 10- integral its and so 10'^; part lies on. in its Hence a between 10 ~^ 10 . Let then N be a number whose integral part contains m digits ^Y_ \Q(n - 1) + a . • . Hence the log N= (« — characteristic is 1) ?i fra<;tiou +a — 1 ; . fraction. that is, the characteristic of înitif is less hjj the logarithm of a number greater than the number of digits in its integral part, and Example. The log 87-263, are resiêctively number To determine less positive. one than characteristics of log 314, (ii) is 2, the 1, log 2-78, 0, log 3500 3. characteristic of the logarithm of a than unity. decimal with one cipher immediately after the decimal and less than '1, lies between 10 ~- and 10 ^; a number with two ciphers after the Hence decimal point lies between 10~^ and 10 -; and so on. A point, ?juch as '0324, being greater than '01 a decimal fraction with n ciphers immediately after the decimal point lies between lO * ^*) and 10 , http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 172/467 8/12/2019 Elementary Trigonometry by Hall and Knight Tifit 143 T.OGARITHMK. XTv] D Ijc <leclmal beginning with .1 logi>= — (m + .•. Hence the characteristic is 1) ii +a ciphers; then fraction. — (71 + I); that is, the characteristic of the logarithm of a number less than one is negative and one more than the number of ciphers immediately after the decimal point. The Example. log characteristics of -4, -li are respectively The 163. maiitiss(e numhers which have log-00013.5, log -3748, the -1, -4, log -08 -2. are the same for the logarithms of all same significant digits. same sequence of digits, differing only in the position of the decimal point, one must be equal to the other multiplied or divided by some integral povfer In Hence their logarithms must differ by an integer. of 10. For if any two numbers liave the other words, their decimal parts or mantissse are the same. Examples, (i) (ii) log 32700 = log (3-27 x 10^)= log 3-27 log -0327 = log 3-27 =log (3-27 x IQ-^) = log 3-27 (iii) 10* + 4. = log 3-27 + log IO-2 -2. log-000327 = log(3-27xl0-') = log 3-27 + log = log3-27 + logl0-^ -4. Thus, log 32700, log -0327, log -000327 differ from log 3-27 only in the integral part; that is the mantissa is the same in each case. Note. tively. integral The characteristics of the logarithms are 4,-2,-4 respecThe foregoing examples shew that by introducing a suitable power of 10, all numbers can be expressed in one standard form in which the decimalpoint ahvaijs stands after the first significant digit, and the characteristics are given by the powers of 10, without using the rules of Art. 162. The logarithms of all integers from 1 to 20000 have In Chambers' ]Mathematical Tables been found and tabulated. they ai-e given to seven places of decimals, but for many prac164. purposes sufficient accui-acy is secured by using four-figure These are avaibible for all numbers from 1 to 9999, logarithm.?. and their use is explained on page 163^. tical http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 173/467 8/12/2019 1 Elementary Trigonometry by Hall and Knight RI.EMKNTARY TRTGOXOMRTRY. 44 [fHAr. Advantages of Common Logarithms. It will now be seen that it is unnecessary to tabulate tlie characteristics, since they can always be written down by inspection [Art. 162]. Also the Tables need only contain the mantissa of the logarithms 165. of integers [Art. 16.3]. In order to secure these advantages it is convenient abrays to keep the mantissa positive, and it is usual to write the minus sign over a negative characteristic and not before it, .soas to indicate that the characteristic alone is negative. Thvis 4-3010.3, which 0002, is equivalent to —4 + -.30103, and must is the logarithm of be distinguished from —4-30103, in which both the integer and decimal are negative. tlie In the course of work we sometimes have to deal with In such a case an aritha logarithm which is wholly negative. metical artifice is necessary in order to write the logarithm with mantissa positive. Thus a result such as —3-69897 may be 166. transformed by subtracting 1 from Thus 1 to the decimal part. integral part tlie and adding -3-69897= -3 -1 + (1- -69897) -30103 = 4-3010.3. = -4 + Example Required the logarithm of -0002432. 1. In the Tables we find that 3859636 is the mantissa of log 2432 (the decimal point as well as the characteristic being omitted) ; and, by Art. 163, the characteristic of the logarithm of the given number is - 4 .-. log -0002432= 4-3859636. Find the cube root of Exiuiiplc 2. log 7 = log 887904 -8450980, Let X be the required cube root log that .T = I log (-0007) = I -0007, ; having given = 5-9483660. then (4-8450980) = J (6 + 2-8450980) = 2-9483660 887904 = 5-9483660; = log X is, log but .-. .T 167. The logarithm from log 2 ; of j -0887904. and its powers can easily be obtained for log .5 = log — = log 1 - log 2 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight = 1- log 2. 174/467 , 8/12/2019 Elementary Trigonometry by Hall and Knight 145 LOGARITHMS. XTV.] EXAMPLES. XIV. a. Find the logarithms respectively 1. of the numbers 1024, 81, -125, '01, -3, 100, 2, ^3, 4, -001, i, -01. to the bases Find the values of 2. 243, log, log, 16, log,9 343^/7. log.oilO, Find the numbers wliose logarithms respectively 3. to the bases -25, 49, - 2, Find the 4. . 1 ,, are -03, -1^ 1 , --, -1, 2, 100, •04, 1, -4. 1-a, i-espective characteristics of the logarithms of 325, 1603, 2400, 10000, 19, 3, 11. 7, 9, 21. to the bases Write down the characteristics of the common logarithms 5. ' of 523-1, 3-26, The mantissa 6. -03, 0002, 1-5, of log 64439 is 3000-1, ^ -1. -8091488, write down the logarithms of -64439, 6443900, -00064439. The logarithm of 32-5 numbers whose logarithms are 7. is 2-5118834, -5118834, = -3010300, log 3 down the 4-5118834. [^When required the foUowing logarithms log 2 write 1-5118834, = -4771213, may log 7 be used = -8450980.] Find the value of 8. log 768. 9. log 2352. 11. logv%04. 12. log 14. log cos 60°. 15. log sin^ 60°. nV -001 62. 10. log 35-28. 13. log -0217. 16. log \^sec 45°. Find the numerical value of ,„ 17. ^, 15 25 „, 4 2log-^-log-2 + 3log-. , http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 175/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 14fi — - 4 log g 18. Evaluate 19. Find the seventh, root of 1 6 log b'ind the ( 7 log . ^ ^ = -1207283. cube root of -00001764, = 5-415499.5. given log 260315 21. - 7, given log 1-320469 20. - [OHAP. find the logarithm of Hven log 3571 =3-5527899, 3-571 X 03571 x\''3571. (liven log 22. 1 1 ^- 1-0413927, find the logarithm of (-00011)3 X (1-2] )-x (13-31)* Find the number of 23. How many the base is 7 12100000. digits in the integral parts of Vl25/ 1,20/ 24. -T- positive integers have oliaracteristic 3 when ? Suppose that we have a table of logarithm.s of numbers to base a and require to find the logarithms to base h. 168. Let N be one of the numbers, then logj Let y = logi N, so that .-. that hv required. is = N. = log^X; log«(62') is, X y\ogah = \ogaX; 1 X log„ 3; loga b or lotr, Now X= , r log„i N X (IV los^„ .\' and h are given, log„ i\' and log,, may be found. from the Tables, and thus log;, since Hence U) base h it N appears that we have only constant quantity and ai'c /> t(j transform logarithms from base a to multiply is known them all by given by the Tables ; . -. it is ; this is known a as the modulus. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 176/467 8/12/2019 Elementary Trigonometry by Hall and Knight LOGARTTHMS. '^TV.] we put a Tf in equation (1) logj. rt = in the in be given. The a future section. Example iog:6xi'-'^''«=i^^ l. necessary logarithms use of four-figure Tables will be explained Given log 2 1. examples following will N, we obtain for log6axlog„& = .-. 169. 147 all = -3010300 and log 4844544 = 0-6852530, fiud the value of (6-4) iVx (^^256)=»-^s/80. Let x be the value of the expression 1 >«^ 3 64 , ; then 256 . 1 ^,^ , = IO^ ^l0 + 4^°îO00-2^ -^ '^ .- ^ (log 2« - 1) +1 (log 2« - 3) - ^ (log 2^^ + 1) /I 9 „ 1\ /6 3\, ^(l0 + ^-2/°^'^-U)-^4+2) ,, = f5 + ~)log2-2it = 1-5051500 + -0301030 - 2-85. Thus log .1; = 2-6852530. log 4844544 = 6-6852530, But .T= -04844544. .-. Example 2. point and the Find how many ciphers there are between the decimal in (-0504)^*^; having given first significant digit log 2 =-301, log 3 Denote the expression by E ; = -477, log 7 = '845. then ^°g^ = i«^°^To-ô = 10 = 10 (log 504 - {log (28x32x7) -4} 4) = 10{31og2 + 21og3-|-log7-4) r^ 10 (2 -702 - 4) = 10 (2-702) 310^2=. -903 2 log 3= log 7 = -845 -954, 2702 = 20+7-02 = 13-02. Thus the number of ciphers is 12. [Art. 162.] H. K. E. T. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 11 177/467 ' 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 148 [OHAP. Exponential equations. 170. If in an equation the unknown quantity appears as an exponent, the solution may be efiected by the help of logarithms. Example Solve the equation 8^ ''*= 12^^-^, having given 1. log 2 From 3 log 3:^ -47712. the given equation, by taking logarithms, we have .-. = -30103, and - (5 8x) log 8 (5 -3a;) log 2 2a;) = (4 = 10721 4log.S=-l-9084S 1987.-i = (4 -2a;) (2 log 2 + log 3); .^j .• = r (9 log ' '^ Thus 5 log 2 X— -3t) Example 55091 2. -55091 Given log 2 log 2 + (/ log 5 = Thus — -30103, (x li x eliminating y, a +b solve the simultaneous equations = 0, r(, (a- log5 = + - b ) log 2 H log 5 1) ; log 5 / log 2 = log 2. = 0, — or hx + nij — a - b. - b (a - b), log 10 ''=7log5r^«j-log2b + /<. + 'i)+ay — a, [a- = 2. 5*+'. 2 ax + by and And ( .Sti '3.S4570 of the given equations For shortness, putlog2 By 198730 nearly. Take logarithms .T ) 1652^3_ 2^.5 = , .-. 95424 log 3) -19873 ^ - 2 log 3 - -.55091' 2-4 log 2-2 _ 71og2-41og3 _ 10^2-1-50515 •21o.e;.S= 81og2-41og3 15 log 2 -2 7 log 2 log 12 -30103. b http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 178/467 8/12/2019 Elementary Trigonometry by Hall and Knight EXAMPLES. [ 149 LOGARITHMS. XI^'.] When required may be the values XIV. of log -2, log b. 3, log *i gufen on jt?. ]4r) uxed^ Find the value of \'.378 X 2. \/I08 -r (v^l008 x ^486), given log a080)5 X 3. .301 824 = .5-4797536. (•24)S X 810, given log 2467266 = 6-3922l60. Daleulate to two decinial places the values of log, 49. 4. Iog2o800. 7. Find how many ciphers there are 5. log,2A-l<>^^- 6. liefore the first significant digits in (•00378)'^' To what base 8. given log 11 is and (•0259)3«. 3 the logarithm of llOOO = 1 -0413927 and log 222398 Solve to two decimal places the equations 9. 12. 13. 2^-1 = 5. 5^= 2 - f and 2=^ =3 and 3'-^ 10. 52-*-*' 2''î = = 7. Given log242 b, = : 11. 51-^ = r, find = 6^-3, -I--'-. Given log 28 = a, log 21=?), log 224 in terms of a, b, r. 15. = 5-3471309. = 3-^-». 14. log 66 in tenns of a, ? rt, log25 log 27 iog80 = /), log45 = r, find log .36 and and c 11-2 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 179/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 150 [OHAP. XTV. MISCELLANEOUS EXAMPLES. Prove that 1. cos (30° +A ) - .4 - cos (60° + J ) cos (30° a = 2, c = v^2, 5=15°, If 4. Shew that 5. If h cos ^4 6. Prove that + tan cos a =a cos B, (1) sin ^ (sin 3^ ,^, sin a + sin -- ^ that Shew XI ^ i. If Z* =« ,,, 9. Shew sin a = cot — sin a - cos ^'os —. is isosceles. (v/3 ,, , - 3a , 1 ), . , sin 3a 1 cos a C= 30°, that tan 4a ^ = z = 2^ cot ^2a. , find .1 and B. —— — + 4tana-4tan3a — —~ -6 .-, tan-* a r— tan* a . In a triangle, shew that 2B + b'^ cos a- cos (2) 4 (he c,os- If .^r^ + //' ^ 2 .4 = a^ + h'^ - 4ab sin A + ca cos- — + ab cos- + «* = 2<;''^(rt^ + ft'-^), - j sin fi = (a + + ft ; r) -. prove that C'=45° or 135°. [Sohê as a quadratic in 12. ; = tan 4a. ^ sm a (1 11. triangle a. 5^ + sin 7^ + sin 9^) = sin 6^ sin4^ 1 10. C . + cos 3a + cos 5a + cos 7a c.i 8. = ^. solve the triangle. ' 7. B . shew that the -^ cos a yi ) 222 A . + sin 3a + sin 5a + sin 7a (2) - that sm2A + 8in2B + sm2C sin ^+sm^ + smc7 3. cos (60° ) 1{A + B + C= 180°, shew 2. E. If in a triangle ct)S 3.4 one angle mu.st be <;'-.] +cos 35 + cos 3(7=1, shew that 120°. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 180/467 8/12/2019 Elementary Trigonometry by Hall and Knight CHAPTER XV. THE USE OF LOGARITHMIC TABLES. Seven-Figure Tables. {The use of Four-Figure Tables toill he found explained on page 163^.) In a book of Seven-Figure Tables there will usually be found the mantissce of the logarithms of all integers from 1 to 100000; the characteristics can be written down by inspection 171. and are therefore omitted. The logarithm of [Art. 162.] any number consisting of not more than 5 significant digits can be obtained directly from these Tables. For instance, suppose the logarithm of 336 34 is required. Opposite to 33634 we find the figures 5267785; this, with the decimal point prefixed, is the mantissa for the logarithms of all numbers whose significant digits are the same as 33634. We have therefore only to prefix the characteristic 2, and we obtain log 336-34 = 2-5267785. log 33634 log -0033634 = 4-5267785, = 3-5267785. Similarly, and 172. Suppose now that we required log 33634-392. Since this number contains more than 5 significant digits it cannot be obtained directly from the tables but it lies between the two consecutive numbers 33634 and 33635, and therefore its logarithm lies between the logarithms of these two numbei-s. ; If 336.34 to 33635, making an increase of 1 in the the coii-esponding increase in the logarithm as obtained we pass from number, from the tables now we pass from 33634 to 33634-392, making an increase of '392 in the number, the increase in the logarithm will be -392 x -0000129, provided that the increivse in the logarithm is proportional to the increase in the is -0000129. If number. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 181/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TKIUUNOMKTUY. 152 [chap. 2 4J <>9 W — t (N « W5 CC '* « 1^ o; a w ^ 1^ CO c. O IM (N rH 05 C5 oo t- >« lO CO rH -# <M CO -* CO t- O CO OS 00 -I COl« O IM O 55 CC T-H I Si,-< O i-l fO «5 05 00 C^ CO >0 00 T-H -^ >o 00 o O o ^ fee ;3 lO iM -^ lO -Tfl I CO (M r-l CO 03 rH CO CO 05 0-] CO * >0 t^ I >0 >0) <1< ec CO C5| O o IM O ^ o O I * LO CO C5 OO _u ^' (M -^ C^ lo CO t~ I t>-l O ^ * O <M tr~ CO -^ lOl CO (M -H Ct CO CC 05 OJI'* lO CO t- Oslo I— ^ Tjl o o CO ^ >0 -f CO CT t^ OO 00 >-•:> O 00 CO Ol T-H t^ CO l.O -<J< CO' >-( t~ O' CO 01 CO CO t^ 1 tCO CO Oslo O CO (N CO t^ 05 O O] -:*< i-H CO >»l l^ IC t^ CO CO CO C^ t~ rH CO CO CO CO -^ CO 05 lO CO l^ t^ t^ >0 CO CO C5 <M I— (M T^ CO -^ -M i-H »0 00 lO •-H O C-. ^ I O to CO lO * O 00 00 00 L^ CO 05 1-1 OS 00 >.0 00 1—1 CO CO T-l IM 1* IC CO O 2 <ii J '^ • ^ ^ 3 — ^ , • g =* 1^ S -« «« £ tc ~ i SO ' 'E .2 . -* lO l^ 00 C5 -^ CO iM r-t OS c^ lo 00 K-H CO t- CO I ,,3 '-^ ' i 5 g o^'fl « cj CO ;:5 p 03 -^ gcS OJ ^ ^ ^ CO O X th CO '^^ I 00 Ci CO t^ 05 00 CO 00 o 'o t^ OS e-, 00 t^ CD CO *— ' CD 05 (M 110 »*< >0 t- 00 O o o op cri C5 i-i lO 00 05 05 05 05 00 05 CO C~ CO CO CO 05 IM 1-1 (M CO lO l^ >o lO CO OO 05 00 t~ CO kO CO C<I lO 01 CO t-l -^ i-l C^ CO to CO O O rH tH CO cd i-i -^ t~ iH C<I CO lO 1-1 -it ^ CO CO += OHO js C5 05 iM CO ^ :a) http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 182/467 8/12/2019 Elementary Trigonometry by Hall and Knight SEVEN-FIGURE TAHLES. XV.] log 33651 =45269980, log 33652 = 4-5270109, Similarly, and 153 the transition in the mautissse from 526... to 527... being shewn by the bar drawn over 0109. This bar is repeated over each of the subsequent logarithms as far as the end of the line, and in the next line the mantissse begin with 527. Example. From Find log 33634-392. the Tables, log 33635 = 4-5267914 log 33634 ^4-5267785 difference for 1= •0000129 0000129 Now by the Rule of Proportional Parts, log 33634-392 will be greater than log 33634 by -392 times the difference for 1 hence to 7 places of decimals, we have •392 •258 61 ; log 33634 proportional difference for .-. 7 568 =4^5267785 ^392= •0000051 log 33634-392 = 4-5267836 without the number, on multiplying by -392 we obtain the product 50-568, and we take the digits given by its integral part (51 approximately) as the propor- In practice, the difference for 1 ciphers; if therefore we treat is usually quoted the difference 129 as a luhole tional increase for -392. 175. for -392 The method of calculating the proportional difference which we have explained is that which must be adopted when we have nothing given but the logarithms of two connumbers between which lies the number whose logarithm we are seeking. secutive But when the Tables are used the calculation is facilitated by means of the proportional differences standing in the column to the right. The This gives the differences for tenths of iniity. difference for -392 is obtained as follows. •392xl29 = (A + ^^+_|_)xl29 = 39-Ml-6-F -26 = 50-86. The difference for 9 quoted in the margin (really 9 tenths) is 116, and therefore the difference for 9 hundredths is 11-6; and similarly the difference for 2 thousandths http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight is -26. 183/467 , 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 154 [chap. In practical wurk, the following arrangement =4-5267785 log 33634 add 39 3 for 116 9 26 2 log 33634 -392 .-. 176. The adopted, is = 4-5267836 following example solved more concisely as a is In the column on the left we work from in the column on the right we obtain the data of the question the use of the Tables independently of the two the logarithm by model for the student. ; given logarithms. Find Example. log 33 -656208, having given log 33656 = 4-5270625 = = -001 = for From 1-5270754 1-5270625 log 33-657 log 33-656 diff. and log 33657 = 4 -5270754. the Tables, =1-5270625 log 33-656 129 •208 add for we have 26 2 11032 103 8 2580 log -000208= 26 832 =1-5270625 log 33-656 diff. for 33-656208= 1-5270652 log 33-656208 = 1-5270652 The Rule of Proportional Parts also enables us the number corresponding to a given logarithm. 177. to find Find the number whose logarithm is 2 5274023, having given log 3-3683 = -5274108 and log 3 -3682 = -5273979. Example 1. Let X be the required number ; then log .T= 2-5274023 = 2-5273979 diff.= 44 log -033682 that is =2-5274108 log -033682 = 2-5273979 -000001= 129 for between -033682 and -033683, 44 greater than -033682 by ^^ x -000001 hence x and diff. log -033683 is lies 129 129 ) 440 ( 34 387^ 530 51(i by -00000034. .-. x= -03368234. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 184/467 8/12/2019 Elementary Trigonometry by Hall and Knight 155 SEVEN-FIGURE TABLES. XV.] lu working from the Tables, we proceed as follows, log :^ a; 2-5274023 =2 •5278979 log -033682 44 39 3 50 .x = -03368234. of 129 are saved the trouble of the division, as the multiples which occur during the work are given in the approximate forms 39 difference column opposite to the numbers 3 and 4. and 52 in the We Example 2. Find the fifth root of log 2-5612 =-4084435, log 3-0317 Let a; = (-0025612)°; -0025612, having given = -4816862, log 3 -0318 = -4817005. then log ..-=- log (-0025612) =i {3-4084435) =i (5 +2-4084435) = 1-4816887. .-. =1-4816887 log log -30317 = 1-4816862 log -30317 diff.= 25 proportional increase= Thus -30318= logo; .1 = 25 j^ diff. x -00001 = for -00001 = -00000175. 1 -4817005 1-4816862 143 US ) 250 ( 175 1^ 1070 1001 -30317175. 690 715 EXAMPLES. XV. a. Find the value of log 4951634, given that log 49517 = 4-6947543. log 49516 = 4-6947456, Find log 3-4713026, having given that log 34714 = 4-540.5047. log 347-13 = 2-5404921, Find log 2849614, having given that log 2-8496 = -4547839, log 2-8497 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight = -4547991. 185/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 156 [cHAP. Find log57 63325, having given that 4. log 576-33 = 2-7606712, Given log 60814 = 4-7840036, 5. = 3-7606788. log 5763-4 difF. for 1 = 72, find log 6081465. Find the number whose logai-ithm 6. log 55740 = 4-7461670, 4-7461735, given is = 4-7461748. log 55741 Find the number whose logarithm is 2-8283676, given log 67355 = 4-8283698. log 6-7354 = -8283634, 7. Find the number whose logarithm 8. log 1068-6 = 3-0288152, log 1-0687 Find the number whose logarithm 9. log 8-2877 = -9184340, log 8287-8 Given log 253-19 = 2^4034465, 10. number whose logarithm is 1 and = -0288558. 3-9184377, given is = 3-9184392. diflf. for 1 = 172, find the -4034508. Given log 2-0313 = -3077741, 11. 2-0288435, given is log 1-4271 log 2-0314 = -3077954, = -1544544, find the seventh root of 142-71. Find the eighth root of 13 89492, given 12. log 13894 = 4-1428273, log 138-95 = 2-1428586. Find the value of v^242447, given 13. log 2-4244 =3846043, diff. for 1 = 179. Find the twentieth root of 2069138, given 14. log 20691 = 4-3157815, diflF. for 1 = 210. Tables of Natural and Logarithmic Functions. 178. Tables have been constructed giving the values of the functions all angles 0° 90° at between and sines, These ai-e called the Tables of natural In the smaller Tables, such as Chambers', cosines, tangents,... trigonometrical intervals of 10 . the interval is 1'. The logarithms Since many of of the functions have also been calculated. of the trigonometrical functions are less than luiity http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 186/467 8/12/2019 Elementary Trigonometry by Hall and Knight 157 SEVEN-FIGURE TABLES. XV.] are logarithms are negative, and as the characteristics To be omitted. not always evident on inspection they cannot the characteravoid the inconvenience of printing the bars over are all increased by 10 and are then istics, the logarithms under the name of tabular logarithmic Sines, registered their cosines,. •• The notation used is LcosA, Lta,ii6 L sin A = log sin A + thus ; 10. For instance, L sin 45 ' = 10-1- log sin 1 45 =104- log -j^ = 10 - - log 2 = 9-8494850. With 179. noticed, certain exceptions that need not be here natural sines, cothe rule of proportional parts holds for the logarithmic sines, of all angles, and also for their sines,... that as In applying this rule it must be remembered cosines,.... 0° to 90° the functions sine, tangent, the angle increases from cotangent, cosecant secant increase, while the co-functions cosine, decrease. Example From 1 . Find the vahie of sin 29° 37' 42 . sin 29° 38'= -4944476 the Tables, siu 29° 37' =-4941948 diff. for .-. = prop' increase for 42 J~ 60 = 2528 = 1770 x 2528 sin 29° 37' .-. Example Let A 2. sin 29° 37' 42 = ; cos A 60 is less cos 16'; hence A ^y ^ ^° ' *^ * Thus the angle is '' 43 ^y '^ = '7280843 J 873 60 prop' part must be greater than 43° 16' 1^9?4 -7280843. cos 43° 16' = -7281716 1994* than cos 43° is then from the Tables, cos 43° 16' = •7281716 cos 43° 17' = -7279722 But -4943718. Find the angle whose cosine be the required angle diff. for = -4941948 ' '^^- I ^yg^ ^ 52380 ( 26 3988 n964 16' 26 . http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 187/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 158 [chap. lu order to illustrate the use of the tabular logarithmic functions we give the following extract fi'om the table of logarithmic sines, cosines,... in Chambers' Mathematical Tables. 180. 27 Deg. Sine Diff. Cosec. Secant 9-6570468 9-6572946 9-6575423 9-6577898 9-6580371 2478 2477 2475 2473 10-3429532 10-3427054 10-3424577 10-3422102 10-3419629 10-0501191 9-6706576 9-6708958 9-6711338 9-6713716 9-6716093 2382 2380 2378 2377 10-3293424 10-3291042 10-3288662 10-3286284 10-0537968 10-0538638 10-0539308 Cosine Diff. 10-0501835 10-0502479 10-0503124 10-0503770 10-3283907 10-0539979 10-0540651 Secant Cosec. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight I). Cosine 644 644 645 646 670 670 671 672 D. 188/467 8/12/2019 Elementary Trigonometry by Hall and Knight SEVEN-FIGURE TABLES. XV.] Example From Find 1. /> cos 02° 57' 12 . L cos 62° 57' = 9 -6577898 L cos 62° 58' = 9 -6575423 the Tables, diff. for .-. Example 2. A when E =— - bO L = 9-657740H cos 62° 57' 12 L A= T. 495 sec 27° 39' = 10-0526648, yl 3' sin 38° 4' 2. 35 , 24' 37-5 , Find cosec 55° cosec 55° 22' 4. 5. 6. 7. 21' 28 , . XV. b. 3' = -6163489. tan 38° 24' = -7925902. having given that = 2-1815435, cosec 55° 21' = 1-2155978. = -8600007, is 2-1809460, given sec 62° 42' Find the angle whose cosine is cos = -8764620, = 2-1803139. -8600931, given .30° Find the angle whose cotangent cot 48° 46' = 55 having given that = 1-2153535, cos 30° 41' 10 . sin 38° Find the angle whose secant sec 62° 43' = 1]0, having given that tan 38° 25' = -7930640, 3. -— x = 27°39'55 = -6165780, Find tan 38° ]() 605 EXAMPLES. sin 38 for ^ proportional increase = Thus Find diff. = 10-0527253 sec 27° 39' = 10-0526648 sec diff. 1. = 495. 10-0527253. L .-. x 2475 = 9-6577898 Given sec 2475 12 62° 57' Subtract for 12 Loos find 60 proportional decrease for 12 .-. 159 40' is -8766003, given cot 48° 45' L sin 44° 17' 33 , given /;sin44° 18' = 9-8441137, 7:sin44° = -8601491. = -8769765. Find http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 17'=9-8439842. 189/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 160 L Find 8. cot 36° 26' 16 , given /.cot 36° 27' Find 9. = 10-1315840, L cos 55° X cot 36° L cos 55° 30' the angle whose tabular the data of example 7. using 9-8440018, Find 10. 26' = 10-1318483. 30' 24 , given 55° 31' =9-7529442, Zcos [CHAP. = 9-7531280. logarithmic whose tabular logarithmic 9-7530075, using the data of example 9. 11. Find 12. Given L tan 24° the angle 50' = 9-6653662, difi: for 1' sine is cosine - 331 3, is find 50' 52-5 . 7vtan24° (liven 13. L cosec 40° 5' = 10-1911808, Z;cosec40° 4' diff. for 1' = 1502, find 17-5 . Considerable practice in the use of logarithmic Tables will be required before the quickness and accuracy necessary in Experience shews all practical calculations can be attained. 182. that mistakes frequently arise from incorrect quotation from the Tables, and from clumsy arrangement. The student is reminded that care in taking out the logarithms from the Tables is of the importance, and that in the course of the work he should learn to leave out all needless steps, making his solutions as first concise as possible consistent with accuracy. Example From Divide 6-64'2o69H by -3873007 1. the Tables, = log 6-6425 =1-5880475 log -38730 •8228316 7 40 6 5 9 log 7 -3873007=1 '5880483 3 log 6-6425693 log -: J8223362 -3873007-1 -5880483 By subtraction, we obtain From the Tables, log 17-150 1-2342879 - 1-2342641 *^38 229 9 :^ ThiiH tliP quotient is 90 70 17-15093. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 190/467 8/12/2019 Elementary Trigonometry by Hall and Knight 161 SKVEN-FIGURE TABLES. XV.] Example of a right-angled triangle is 3-141024 side is 2-o93167; find the other side. and one Let The hypotenuse 2. c be the hypotenuse, a the given then From 2 and .r the side required; c=3-l4l024 = (? - a-={c+ a)(c - a) log x = log (c + a)+ log (t'-rt) x'^ .-. side, a = 2-593167 ; the Tables, <;+a = 5-734191 C'fl- 047858 •7584653 log 5-7341 68 9 1 1-7386617 55 log -fAlHr, By Dividing by 2, 4971394 addition. we have •2485697 -2485617 l0g.T log 1-7724 '80 ZL 60 49 Thus the required side is 1-772432. EXAMPLES. [/yt Figure XV. this Exercise the logarithms are c. to he taken Seven- Tables.'] 1. Multiply 300-261^ bv -0078915194. 2. Find the product of 23.5-6783 and 357-8438. 3. Find the continued pi'oduft of 153-2419, 2^8(i32503, 4. from Divid.' 1-0304051 l.y Divide 357-8364 hy and 07583646. 27-093524. -00318973. 5. 6. Find X fi'uiii the equation ()178345.rr- 21 -85632. 7. Find tlio value of 3-78956 X 0536872 -V -(MJ7291 6. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 191/467 8/12/2019 Elementary Trigonometry by Hall and Knight ET,EMRNTARY TRIGONOMKTRY. 162 8. Find the cube of -83410039. 9. Find the 10. fifth root of 15063-018. Evaluate \/384-731 and \^ 15-7324. Find the product of the square root of 11. cube root of 353246. 12. Subtract [rHAP. the square of -7503269 034-3963 and the 1 from the square of 1-035627. 13. Find the value of (34-7326)^ X \/2^894 >v/4-39682 Example 3. Find a third proportional cube of -3172564 to the and the cube root of 23-32873. Let .1- be the required third proportional (-3172564)3 : (23-32873)* x = (23-32873) whence log .-. From .r == I » ; then = (23-32873) » : .r ; ~ (-3172564)3 log 23-32873 - 3 log -3172564. the Tables, log -31725 -- 5 4 130 7 82 6 1-3678775 log 23-328 1-5014016 5 1-3678910 3 ,T5014103 2 3^ 2-7357821 3 2 -5042811 I -9119274 2-5042311 By subtraction, = 2-4076963 = 2-4076798 log.i; log 255-67 165 153 '120 119 Thus the third proportional is 255-6797. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 192/467 8/12/2019 Elementary Trigonometry by Hall and Knight THE USE OF LOGARITHMIC TABLES. XV.] 14. 163 Find a mean proportional between •0037258169 and -56301078. 15. Find a third proportional to the square of '43607528 and the square root of 03751786. 16. Find a fourth proportional to 56712-43, 29-302564, 17. -.33025107. Find the geometric mean between (-035689)^ and (2-879432)^. 18. Find a fourth proportional to \/32-7812, 19. Find the value of sin 27' 13' 12 20. If 6 : = sin A X sec 112 13' 5 . x cot 47° x sec 42° 53' 15 15' 30 . 6 = 417-2431, 6 = 113° 14' 16 '. sin B, find a, given : ^ = 3.5° 1.5' 33 , 5=119° 14' 18 . Find the smallest values of S which satisfy the equations Find tau'5<9 .' = :^; (2) 3 .sin- ^-|- 2 sin ^= I. from the equation .<xsec28° 26. 20 = 378-25, (1) 25. 1.3' = 3241368, (/ 6 24. 15 . Kind the value of ab sin C. when <f 23. 2' Evaluate sin 20' 22. x cos 46° Find the value of cot 97° 14' 16 21. \/7836-43. ^/357-814, 17' 25 = sin 23° IS' 5 x cot .38^ 15' 13 . Find 6 from the equation sin-^ where a = 32° 47' and |8 ^ = cos- a cot = 41° /a, 19'. 12 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 193/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 163. [CHAP. Use of Four-Figure Tables. 182 To find \. logarithm of a given nnmher the from the Tables^ Example We Find log 1. 38, log 380, log -0038. number 38 in the left hand column on page 374. This, with the decimal Opposite to this we find the digits 5798. point prefixed, is the mantissa for the logarithms of all numbers find the first whose significant we have log 38 Example = 1-5798, 2. The same Hence, prefixing the characteristics digits are 38. log 380 Find log 3-86, log -0038 = 3-5798. = 2-5798, log -0386, log 386000. mantissa of the logarithms From this line we choose the of all numbers which begin with 38. mantissa which stands in the column headed 6. This gives •5866 as the mantissa for all numbers whose significant digits are 386. Hence, prefixing the characteristics, we have line as before will give the log 3-86 = -5866, log -0386 = 2-5866, log 386000 = 5-5866. Similarly the logarithm of any 182b number consisting of not more than 3 significant digits can be obtained directly from When the number has 4 significant digits, use is the Tables. made of the principle that when the difterence between two numbers is small compared with either of them, the difference between their logaritlims is very nearly proportional to the It would be out of place to difference between the numbers. It will be attempt any demonstration of tlie principle here. sufficient to point out that differences in the logarithms corresponding to small difi'erences in the numbers have been and are printed ready for use in the difference cohcmns The way in which these at the right hand of the Tables. differences are used is shewn in the following example. calculated, Example. Find (i) log 3-864 ; (ii) log -003868. Here, as before, we can find the mantissa for the sequence of This has to be corrected by the addition of the figures digits 386. which stand underneath 4 and 8 respectively in the difference columns log 3-86 (i) 4 5 log 3-864 = -5871 diff. for .-. Note. difference =--5866 After a little =3-5866 log -00386 (ii) difif. .-. for 8 9 log -003868= 3-5875 practice the necessary 'correction' from the cohuuns can be performed mentally. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 194/467 8/12/2019 Elementary Trigonometry by Hall and Knight FOUR-FUiURK TABLES. XV.] • 1(53 The number corresponding to a given logarithm is called its antilogaritlim. Thus in the last example 3*864 and •003868 are respectively the numbers whose logarithms are '587 182i', and 3-5875. Hence antilog •5871 To find 182o, = 3-864; antilog 3-5875 the antilogarithm = -003868. of a given logarithm. In using the Tables of antilogarithms on pages 376, 377, it is important to remember that we are seeking numbers corresponding to given logarithms. Thus in the left hand colunui we have the first two digits of the given mantissa;^ with the decimal jjoint prefixed. The characteristics of the given logarithms will fix the position of the decimal point in the numbers taken from the Tables. Example (i) We Find the antilogarithm of 1. 1-583; (ii) in the left 3. Thus -583 is the mantissa of the logarithm of a significant digits are 3828. by Hence 2-8249. hand column and pass along the and take the number in the vertical column headed first find -58 horizontal line (i) antilog 1-583 number whose = 38-28. (ii) antilog 2-824 diff. for .-. == -06668 14 9 antilog 2-8249 ==-06682 Here corresponding to the first 3 digits of the mantissa we find the sequence of digits 6668, and the decimal point is inserted in the corresponding to the characteristic 2. To the number so found we add 14 from the difference column headed 9, i^lacing it under the two last digits of the given mantissa. Ijosition Example 2. Find the product of 72-38 and =1-8591 log 72-3 diff. for 8 for log product Thus 5 =1-7543 log-568 diff, -5689. 9 7 diff. is for 6 6 antilog 1-6146 = 41-17 =1-6146 the required product =41-11 antilog 1-614 41-17. 12—2 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 195/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 103^. ,,. ., , J.xamplc 6. . ,, , , the value of Jbind [CHAP. ••. ... 3-274 X -0059 .-.-Xo Tat? ^ *^^^ Kiguincaiu . digits. By Art. 157, log fraction — log numerator -log denominator. Denominator. Numerator. = log 3-27 4 5 _ log -0059 ^3-7709 diff. for log JiMntt'/'a<oJ- log denomi7iator = antilog 2-228 log /rac<to)i = 2-2282 Example 4. =-01G90 1 antilog 2-2282^-0169] „_^- 3-274 X -0059 Tu '0577 2 for diff. 9 _ =2-8865 log '077 2-2859 -0577 3 for diff. = 2-2859 subtract =11703 log 14-8 -5145 Find the value of .^^r- „, to the Jiearest integer. 4/22 X 6-9 Denote the expression by then .r, log .(=4 (log 330 - log 49) log 330= 2-5185 ^ 1-6902 log 49 - .. (log 22 + log 6-9), = 1-3424 log 22 log 6-9= -8388 •8283 3 |2yi8i2 4 -7271 3-3132 -7271 subtract log .'. = 2-5861 = antilog 385-6, from the Tables. .(=386, to the nearest integer. .r EXAMPLES. [Aitsircrs I'iud \ l)y to be given means of tlie io XV. d. four .Îgmfivant Tables the value Jiyurcs.'] of the following products 1. 4. 2834x17-62. 3-7 X 8-9 X -023. 2. 5. 8-034x1893. 31-9x1-51x9-7. 3. -00567 x -0297. 6. 43 x 807 x -0392. Find tho value of 2-035 ll'i 294-8 * 837-6' 4s7 -:^179 * -08973' http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight ' 6398' 196/467 8/12/2019 Elementary Trigonometry by Hall and Knight XV. FOUR-FTftURE TA}?LES. 1 1(1?^ ii 1472x3805 2-38 X 3-90 1 ^^' , ^^- f83^ '^HT9 • 15-38 X -0137 . 276 X -0038 T ' ^^' • 2-31 x -037 x . 925-9xlo»7 7Tî3~ .. • 1 -43 -0561 x 3-87 x -0091 ' 16. N^5T. 17. i'u. 18. 4^82^. 19. 4'lOl5. 20. (-097)^. 21. (2-.301>\ 22. (51-32)i 23. (-089^7. V— 83x^92 /•0137 X -0296 ^. '*• 873^ na Zb. T? 25. • ii 1 i' b\u(i the value ot 1 ^j—^~ W —^^^t — /•678 X 9-01 ^'^ the nearest, integer. Find a mean proportional between 2-87 and 30()8; and a third proportional to -0238 and 7-805. 27. 28. Find a mean proportional between \^:J47-3 29. By taking find values of .v and \'256-4. logarithms, and solving for log.-p and logy, i/ (to four significant digits) which sati.sfy and the equations Find the value of the expression E.vamjjie 5. ^^'~^-^^ when T/) = 7-8, T=.--, w^-G6, 2f>o \o(i:p log 7-8 r= = logX = log 7r- = 2 logw log 7/- = log -60 jr^ 3-1416, = -8921 -: 1-8195 log 14 =1-1461 log 3-1416 =-9942 2 log -025 \-14, 77 = -0-2r,. =4-7958 -464 1 log -„-^ 2 log 256 =4^164 4-4641 ,,. antilog diff. for antilog 1 =2-911 1 •4W1 =:^»1'> antilog 4-4641^29120. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 197/467 8/12/2019 Elementary Trigonometry by Hall and Knight ETjRMKNTARY trigonometry. 163^, EXAMPLES. 0. Find the value of XV. 27r a When m = 18-34, 32. Calculate the values of 33. where p = 93-75, 34. If/'='—, where ^ = \^î find /^ gr = 33-47, value oi^mv\ = \ '03, w = 4 »- = 9-6, r= 537-6, 35. If s=^ff-, find 38. If .}; y 37. The volume X = 5-875. y TT = 60, ^ = 32-19. 4 V= - irifi, given = 3-1416. / when s = 289-3, = Z\. t 10*, find n when ^ = 73-96 and of a sphere 4 formula y=T..TTr'; find r when Find r from the formula = 8-7 r tlie 35.5 ^jTt;-^, ??i = 3-1416. 7r v = 35-28, find 4 (ii) when 5r= 32-19, 31. p>-», , -^\' Z= 2-863, (i) {Continwd.) d. / 9 [flHAP. tlie of radius r is ?/ = 27-25. given by the radius of a sphere whose volume is 33-87 cu. cm. 38. A 39. If cubical block of metal, each edge of which is 36-4 cm., Find the diameter of the sphere is melted down into a sphere. as correctly as possible with Four-Figure Tables. —= g^^. , calculate '• Also shew approximately tliat v, = 4000, the value of having given that 32-2 g = ^^. 2Trr -r>, —;^, wliere V X 60 X 60 7r = 3-1416, is 24. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 198/467 8/12/2019 Elementary Trigonometry by Hall and Knight yOTTR-FTGURR TABLKS. XV.] Four-Figure Tables of Logaritlunic Functions. 182£. The use Tables of of natural tangents has been explained in Chap. IV. The logarithms sines, cosines, and [See Art. 39^.] of the functions have also been calculated to Since many of the trigonometrical functions are less than unity their logarithms are negative, and as the characteristics are not always evident on inspection they To avoid the inconvenience of printing the cannot be omitted. foiu' places of decimals. bars over the characteristics, the logarithms are increased Viy 10 and are then registered under the name of tabular logarithmic functions. The notation used lent to log sin ^4 is Zsin^4, Xtan^; thus Zsinî is equiva- + 10. — The Tables on pages 384 389 give the logarithmic sines, cosines, and tangents of all angles between 0° and 90° at intervals of 6 minutes. For intermediate angles the difference columns are used in the same way When as for the natural functions. a change in the characteristic in the middle of a horizontal line, the transition is marked by printing a bar over the mantissa. 182f. there is Tlius from the second page of logarithmic tangents we have 84° 85° http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 199/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 200/467 ' 8/12/2019 Elementary Trigonometry by Hall and Knight FOUR-FIGURE TABLES. XV.] EXAMPLES. Find from the Tables the sm2r38'. 48° 46'. sec 4. e. A'ahies of tail 38° 25'. 2. 1. XV. 163„ cot 42° 27'. 5. 15° 11'. cos cosec 55° 22'. 3. 6. Find to the nearest minute the vahies of () from the follow in<T equations sin ^=^-3126. 7. cos 8. <9 = -1782. tan^ = '^. 9, 7 cot^ 10. = -7931. cot^=ir)321. 11. cos(9=|. 12. * Find the values of 13. sin 44° Z; 17'. £ cot 27° 14. 34'. Z cos 15. 72° 15'. Evaluate 16. X cos 46° sin 27 13' s in 34° 17 x tan 82° sin 47° 13' 16'. ' 17. 6' cos 28° . cos l2°3y' Find 20. equation tan the .* = Find the value r-30°16'. na ^ (i) 37° 26' which satisfies aisinC when a = 32-73, of 23. 24. cot-, n when fiivon tan The 26c is ^ h rY -r-sin^^, when (ii) + d) r lengtli cos = <>f = 2-0, r =3 the 6 = 27-86, ti ,sin 0, find „^- the hisectoi- of — d> K). wlien r=-3 ), ^ — tlie angle this, lind its . r, Assuming = 32-78, -^^' </(] 3, = 8: 2e A Fvaluate 7* —— ] 2 ^ c 6 e 4' Calculate the value of 22. 25. angle positive .sec 1 • 21. ABC • smallest ^r, tan 22° 27' c ^^? -.j- value 14'. of a triangle when , = 19-23, -e)2cos^n .1 . )6° , .l-lir) 34'. when <' = 48, a=-23°, </ = ,3219, = -37. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 201/467 8/12/2019 Elementary Trigonometry by Hall and Knight CHAPTER XVI. SOLUTION OF TRIANGLES WITH LOGARITHMS. (The section on Four- Figure Tables, page 183^ inay he taken after Art. The examples on the 183. sohitiou of triangles in Chap. XIII. on the formulae connecting the sides but in practical work much of the arithmetical calculation is avoided by the use of furnish a useful exercise and angle of a triangle labour 188.) of ; logarithms. We shall be used or the formulae of Chap. XIII. may use in connection with logarithmic now shew how adapted for Tables. 184. To find the functions of the half-angles in terms of the sides. We have 2 sin^ — = 1 - cos ^ ^ 26c {b^-2hc-{-c^) 2fcc-6^-c2 26c a-2 - (6 cf + a^ _ a^~ _ + 6-c)(a-6+c) (a 26c 26c 26c a + 6 + c = 2«; Let a+6- c = 2s- 2c= 2 then c (.•,•- c), = 2.9-26 = 2(s-6). a-6 + and „ . ^si 4{s-c){8-b)_ 2i8-b)is-c) ,A ~ 2 . •' be 26c A ^' 2 = /{s^b){g-c) V be—' http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 202/467 8/12/2019 Elementary Trigonometry by Hall and Knight .SOLUTION OP TRIANGLES AVITH LOGARITHMS. Again, 2 cos- — = 1 + cos ^ = 1 H _ (h + cf-a^ _ •ibc ~ ^. { + c-Va){h + c-a) h A = 4s (s — a) 2s 26c Also A = tan ^ VI Similarly B n 2 = V — a) bo (s-b){s-c)^ be s(5-a)' be V 2 •20S s{s tan4=./(Îp:Z). .-. B'_ 6c 2i ^ . a) A — 4- cos A Z ^ sin Z 185. - (s ^ ' 2bc ^211 ¥~ 1(15 may it s(s-a) be proved that / (s-c)(s-a) V ea C_ / (.-a)(s-6) S6 ^'^2~V ' /s{s-b) A / -^ V *^ 2-V COS , ' ca C -= 2 s(.-6) V . . ' /s(s-e) / ^ , a6 ^ ; ' *^ 2-V ~7^^3^- ' In each of these formulae the positive value of the square root must be taken, for each half angle is less than 90°, so that all its functions are positive. 186. To find sui ^4 A sin = 2 sm — in ternui of the sides. cos — s{s-a) {s- b){s-e) -y . . We may sin ^^ be be 2 A = j- also obtain \/s {s tliis — a) {s — b) (s — c). formula in another way which is instructive. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 203/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 204/467 8/12/2019 Elementary Trigonometry by Hall and Knight SOLUTION OK TRIANGLES WITH LOGARITHMS. XVI.] a 187. To From the forinulH solve trionglc when *'^''2-V ^ = log tail whence — may Similarly then G from B - {log {s - h) 167 the three sides are given. ' s{s-a) + log (s - f) - log s - log (*• - a)) ; be obtained by the help of the Tables. can be found from the the equation foi-miila for tan — , and C= 180° — A — B. In the above solution, we shall require to look out from the tables four logarithms only, namely those oi s,s-a, s — b, s~c; whereas if we were to solve from the sine or cosine formulae we should require six logarithms for ; cos A — = 2 — /s(s-a) / -^ '- V . , and cos 00 B -jr 2 = /s(s~h) a / V , ca so that we should have to look out che logarithms of the six quantities s, s — a, s - b, a, b, c. have to be found by the use of the but if one tables it is best to solve from the tangent formulje angle only is required it is immaterial Avhether the sine, cosine, or tangent formula is used. If therefore alL the angles ; In cases where a, solution given logarithms, the t'lioice obtained from certain of formulfe must depend on the has to lie data. Note. We shall always find the angles to the nearest second, so that, on account of the multiplication by 2, the half-angles should be found to the nearest tenth of a second. In Art. 178 we have mentioned that 10 is added to each of the logarithmic functions before they are registered as tabular logarithms but this device is introduced only as a convenience for the purposes of tabulation, and in practice it will be found 188. ; that the work is more expeditious if the tabular logarithms are The 10 should be subtracted mentally in copying not used. down the logarithms. Thus we should write logsin 64 anil in 15'= T-9545793, log cot 18° 3.5' = -4733850, the arrangement of the work care mu.st be taken to keep the mantisste positive. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 205/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 168 Eaample greatest The 1. of sides a are triangle angle; given log 2 = -3010300, L cos 47° 53' = 9-8264910, log 3 diff . [chap. 85, tind the <). 5; 4l», = -4771213, for 60 =1397. Since the angles of a triangle depend only on the ratios of the and not on their actual magnitudes, we may substitute for the Thus in the present case sides any lengths proportional to them. — = is the greatest angle, and = we may take a 5, h 1, c 9: then C sides 21 4=y'-^=V 3 1 V^' log3= -9542426 C log cos — = 1 1 - (2 log 3 log 2 - 1). 2 ) ^ Thus 3010300 16532126 1-8266063 -=1-8266063 log cos log cos 47° 53' =1-8264910 Il53 diff. 1153 .-. proportional decrease =—-^ x 60 .-. = 47 1^ 1153 60 = 49-5 ; 1397 ) 69180 5588 ( 49-5 13300 12573 52' 10-5 . 7270 Thus the greatest angle Example A an- 95° 44' 21 . a = 283, 6 = 317, f = 428, find all the angles 283 317 / (s-b)(s-e) _ / 197 X 86 ~ s(s-a) [s-a) 'V 514x231' V 2 A log tan -^ From If 2. is = 1 - (log 197 + log V 4-28 2 86 - log 514 - log 231 = 2-2944662 86 = 1-9344985 log 514 = 2 -7109631 log 231 = 2 -3636120 4-2289647 5-0745751 5-0745751 log tan :^ ) 514 =s 231 =s- 86=s-c log 197 2 1028 197=s-6 the Tables, log ) 1-1543896 = 1-5771948 log tan 20° 41' = 1-5769585 diff. 2363 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 206/467 8/12/2019 Elementary Trigonometry by Hall and Knight XVI.] But SOLUTION OF TKIANULKS AVITH LOOARITH.MS. diff. for (JO is 169 3822, http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 207/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 170 4. are [cHAP. Find the greatest angle of the triangle 5, 6, = '7781513, 14' = 9 -8890644, in which tlie .sides 7; given log6 Z cos 39° diflf. 1'= 1032. for = 1'75, c = 2'75, find C given log 2, Xtan 32° 18' = 9-8008365, diff. for l' = 2796. 5. If 6. If the sides are 24, 22, 14, find the least angle; given a = 3, Z) ; Z tan 7. 1 7° 33' Find the given log 2, log If given log rt : 6 2, log If « ; 6 : c= 3, : 47' = 9-8355378, 189. 15 log 13 : : 14, find 4, 10, 11; = c' =3 4 : : 2, = solve a for 1' = 1345. the angles ; 7, 26° 33' = 9-6986847, 29° 44' 9-7567587, tan To = 439. 1' the sides are ditf. dift'. for l'=3159, diff. for find the angles dift . 9-4116146, i,tanl4°28'= 52° 14' 10-1 L for 3, L tan L tan 9. diff. when angle gi-eatest L cos 46° 8. = 9-500042, 108395, diff. triangle having ; l' = 2933. given log for 10 for 10 log 3, 870, = = 435. two sides giceti, 2, and the included angle. Let the given parts 1 'e ^, f, sin > I , and B b sin let C G B - sin C _ lb — kc_h-c ain B + am C ~ kb+kc^ b + c sin then 2 cos 2sm . B+C B+C —g, ' tan . sin ' B-C ;r B-C cos— ~~^ J b+c' B-C -2- B+a 2 b-c b+c' http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 208/467 , 8/12/2019 Elementary Trigonometry by Hall and Knight SOLUTION OF TRIANGLES WITH LOGARITHMS. XVI.] B-G B+ C ^ -^ - —^r— = log (6- rj- lug (6 + (•)+ log cot ^ log tan •. from which equation we can find Also By B + C — 90° — A -: , and is addition and subti-action m From .A h-c A ^^. — — = 90 since B+C b-c, 171 a= • , the cq uation B-C therefore known. we obtain B and — —g- ^ si^ -^ -. = log 6 + log sin A - log sin B log « C. ; whence a may be found. of 1. the sides a and h are in the ratio the included angle C is 60°, find A and B\ given Example If log 2:=^ -3010300, tan diff. for l' 7-3 A-B = a-b cot.C - = ;i—5 cot 30 „— -—T a+b 7 + 3 = 2699. 4 =Yfiv/*i; 10 2 log 2 .-. 3, log 3 ^-4771213, Ltan34°42'=:9-8-403776, ^ and to 7 Iogtan^--21og2-l+2log3; = GOiCWOO Îog3=-2:i85607 ^8406207 .. '^ logtan log tan 34° ^' 2 = 1 '8406207 4^= 1-8403776 2431 diff. , . prop'. 2431 „„„ ,,„ mcrease = ^7r^ X bO =54 2431 . 60 , 269y A-B = 34° 42' 54 2 And By A+B 2 ) 145860 13495 ( 54 10910 10796 . =90^-2 = 60^ addition, and by subtraction, .1 = 94° 42' 54 U = 25°17'6 , . H. K. K. T. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 13 209/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTAKY TRIGONOMETRY. 172 Example If 2. (t = 681, 3 = 50 c=:243, [chap. solve the triangle, by i-2', the use of Tables. ^ tan A-C -—— = a-c cot^B - 438 cot,,,.ooi' 2o° 21' a 2 .-. log tan +c rr- jr^r-, 2 924 log 438^2-6414741 log cot 25° 21'= -3244362 '^^ = log 438 - log 924 2-9659103 25° 21' + log cot log log tan .-. ^^=-0002383 log tan 45° .'. diff. •0002383 = -0000000 diff. And 2383 for 60 is 238;^ 2527 60 2383 prop^ increase = ^^-= X 60 A-C By log fc = 109° 39' 57 , 19° 38' 3 . log sin 19° 38' log 6 ^x3540= log sin 19° 38' 3 50° 42' 38' 3 190. = 1 '5263564 = 2-3856063 log sin 50° 42' = 1-8886513 = 2-7479012 =2-7479010 2-2742576 6 = 559-63. Thus A = 177 log 243 log sin 19° 38' 3 .-. = 1*5263387 - log sin C - log sin 19° .-. 56-6 ' — log = log 243 + log sin log 559-63 ( 14680 c sin .-. 142980 12635 16630 15162 C= subtraction, Again, ) . .J B ~ sin C c + log sin 5 2527 ; 39'. addition, and by = 57 = 45°0'57 ^^-=90^^-1 = 64° Also 924=2-9656720 = 1-5263564 2-7479012 109° 39' 57 , G'= 19^ 38' 3 , b = 559-63. From the formul; tan B-C — — = b-c - 2 , b+c Â cot http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight TT 2' , 210/467 . . 8/12/2019 Elementary Trigonometry by Hall and Knight SOLUTIOX OF TRIANGLES WITH LOGARITHMS. XVI.] it will be seen that if b, c, and B— C are 173 known A can be found ; the triangle can be solved when the given parts are two sides and the difference of the angles opposite to them. that is, EXAMPLES. a = 9, If 1 Ztan If 2. rt b 19° c. = 6, C= 60°, find A and B given log 2, log 3, 6' = 9-5394287, Xtan 19° 7' = 9-5398371. ; = l, c = 9, ^ = 65°, Z cot 32° 30' find A and C ; given = 10-1958127, 28' ; diflF. = 27, f = 23, .1 =44° 30', find B Z cot 22° 15' = 10-3881591, Ztanir 3'= 9-2906713, for dift'. If c 5. given log = 210, a= 110, 5=34° Two 6. If (f = 327, If 5 = 4c, ; -4 : = 6711. C and .1 3 and include an angk given log 13' = 9-63240, 2, = -8512583, =65°, find 30' B for diff. log 5-83 and = diff. C ; 1 ' = 35. A and C 28', find ; given 7656686, for l' = 5859. given log 2, log 3, = 10-1958127, Z tan 43° 1 8' = 9-9742 1 33, If « = 23031, 6 = 7677, ( =30° 1< 2, = 10-505 1500. 46'= 10-2700705, 46'= 9-3552267, Z tan 57° 9. given log 15'= 10-23420, c=256, 5=56° Z tan 61° Z tan 12° given 1.5 find the other angles : log 7-1 8. 17° 21' sides of a triangle are as 5 Z cot 30° Z tan 23° 7. 42' 30 , find 1' ; 2, Z cot of 60° 30' C and If 6 4. log 2, = 10-0988763, difF. for 1' = 2592. C=60°, find A and B given log 2, log 3, 49' = 9-8583357, for 10 = 2662. X tan 51° If 17a = 76, L tan 35° 3. XVI. diff. for 1 ' 10' 5 , find = 253 1 A and U ; tg 2, Z tan 15° Z cot or = 9-4305727, 41' = 9-7314436, 5' diflT. for 10 diff. for 10 = 838, = 501. 13-2 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 211/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight 174 To 191. solve ELEMENTARY TRIG0K03IETRY. [chap. a triangle having given two angles and a side. Let the given parts be denoted by B, C, a then the third angle A is found from the equation A =180° — 5— C, ; . Avhence b may . log 2 be obtained from the equation log c = log a + log sin C— log sin A If ?>=:1000, A=i:o°, = -8010300, B = 180° - 45° Ihe ^ , log 7-6986 (7 = 68^ ., least siae = = hînA log « 17' 40 , find the least side, = -8864118, diff. = 66° 42' 20 68° 17' 40 for 1 diff. = 3 + log -^ - for l'=544. . 1000 sin 45° B sm 66° 42' 20 log sin 66° 42' 20 log sin 66° 42'= 1-9630538 20 = 3-.^ log 2 -log = 3- = 57, rt sin .-. ; may L sin 66° 42'= 9-9630538, „, J - log sin be found. Similarly, c Example. having given = log a + log sin B log b • sin B sm^ a , nd sin 66° 42' 20 . ^'^^* ^ ^ 60 -1135869 5 log 2= -1505150 •1135869 0=2-8864131 log 769-86=2^8864118 .-. log 13 diff. prop', increase Thus the least side is 769-8622. EXAMPLES. 1. U B = m' L = r;r = -22. 15', sin 54° 30' 7>.sin65° 15' C^bA XVLd. 30', (i=- 100, tind c = 9-9 106860, = 99581543. ; log 8-9646162 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight given = '9525317, 212/467 8/12/2019 Elementary Trigonometry by Hall and Knight SOLUTION OF TRIANGLES WITH LOGARITHMS. XVI.] 175 = 55°, ^ = 65°, c = 270, find a given log 2, log 3, X sin 55° = 9-9133645, log 25538 = 4-4071869, log 25539 = 4-4072039. 2. If .4 3. If .4 ; =45° C=62° 41', log 9-2788 = -96749, Zsin 4. 72° 6 ; = 70°30', C=78°10', a = 102, find h and e log 1 -02 = -009, log 1 -85 = -267, log 1 -92 = -283, L sin 70° 30' = 9-974, L sin 78° 10' = 9-990, 20' = 9-716. If 7i If Xsin31° a=123, ^ = 29° 5. = 100, find c given L sin 62° 5' = 9-94627, 14' = 9-97878. 5', (7=135°, find 17', c given log ; ; given 2, = 2-0899051, Zsin 15° 43' = 9-4327777, i>=135. log 32110 = 4-5066403, log If 6. 123 ^ = 44°, Zsin Zsin (7=70°, 6 = 1006-62, find a and c = 9-8417713, 66° = 9-9607302, 44° log 100662 log 103543 ; given =5-0028656. =5-0151212, 70° = 9-9729858, log 7654321 = 6-8839067. If a = 1652, 5 = 26° 30', C=47° 15', find 6 and c 7. Zsin 73° 45' = 9-9822938, log 1-652 =-2180100, Zsin 26° 30' = 9-6495274, log 7-6780 = -8852481, I> = 57, Zsin47° 15' = 9-8658868, log 1-2636 = -1016096, /> = 344. L sin ; To solve a triangle when two one of them are given. sides 192. to Let a, b, A B may then C \h 5 = -sin found from the equation (^'=180° — A A, wo h;ive ; J - /?. sin a C ^=^T' . .-. If the angle opposite be found Agam, . sin B = log 6 — log a + log sin log sin whence Then from be given. and a<b, and logc = log<7 A is + logsin (' B A. ambiguous and there supplementary to each other, and also acute the solution will be two values of two values of (7;ind r. — log sin is [Art. 147.] http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 213/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 176 Example. If 6 = 63, c = 36, C = 29° log 2 .sin B =- = -3010300, 23' 15 , find [CHAV. B = -8450980. log 7 sin 29° 23' = 9-6907721, diff. for 1' = 2243, L sin 59° 10' = 9-9388222, diff. for l' = 7o5. sin C =-;; Sb C .sin log sin 29° 23' 15 sin 29° 23' 15 ; 60 4 .-. log sin = 1 '6907721 x2243 = log = log 7-2 /i given L c =^ ; log 2 61 7= - •5359262 29° 23' + log sin .-. 15 ; 2 log 2 5 = 1-9338662 10' = 1-9338222 = J6020600 log sin log sin 59° 1-9338662 440 diff. prop', increase 440 = 755 5 = 59° .-. 8450980 440 x60 = 35 60^ ; 755 10' 35 . ) 26400 2265 ( 3y 3750 value another the above, to is c < b there Also since supplementary of B namely B = 120° 3775 49' 25 . EXAMPLES. XVI. e. tx1. If a= 145, 6 = 178, 7? = 4r A; given find 10', 10' log 178 log 145 2. 3. If 2-2504200, = = 2-1613680, A = 26° 26', b= 121, a log 1-27 = -1038037, log 8-5 =-9294189, If c = 5, 6 = 4, L X sin 32° .sin C=45°, = 85, Z X 9-8183919, = = 9-7293399. 41° 25' 35 find B sin 26° 26' sin 41° 41' 28 find given ; AandB; = 9-6485124, = 9-8228972. given = -30103, L sin 34° 26' = 9-7525750. = 1405, 6 = 1 706, A = 40°, find B given log 1 -706 = -2319790, log -405 = -1476763, log 2 4. I f ; 1 /, sin 40 = 9-8080675, .lift', L sin 51° 1 8' =9-8923342, for r-.-1012. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 214/467 8/12/2019 Elementary Trigonometry by Hall and Knight SOLUTION OF TRIANGLES WITH LOGARITHJfS. XVI.] 5. If ^ = 112° 4', log 573 L sin 39° 6 = 573, c = 394, ^ find 177 and C; given = 2-7581546, log 394 = 2-5954962, 35' = 9-8042757, = 1527, difi . for 60 Zees 22° 4' = 9-9669614. = 8-4, c-= 12, ^ = 37= 36', find A given log 7 = -8450980, L sin 37° 36' = 9-7854332, Z sin 60° 39' = 9-9403381, diff. for l' = 711. 6. If 6 7. Supposing the data ; for the solution of a triangle to be in the three following cases, point out whether the solution a.s will be ambiguous or not, and find the third side in the obtuse angled tiiangle in the (i) ambiguous case A = ZO\ a = 125 feet, ^=30°, a = 200 feet, = 250 c = 250 = 30°, a = 200 feet, c= : (ii) J. (iii) <• 125 feet, feet, feet. Given log 2, = -7809578, = log 6-0390 -7809650, L X log 6-0389 Some sin 38° 41' = 9-7958800, sin 8° = 91789001. 41' which are not primarily suitable for working with logarithms may be adapted to such work by 193. formulae various artifices. 194. To adapt the formula c-^cfi-^l)^ to logarithmic tation. We c-=a^ have W ( 1 +— «' V Since an angle can always be found whose tangent to a given numerical quantity, obtain c2 = (,2 ( _j_ .-. . The angle 6 is - . log c compu- we may put - = tan tan- &) c=a = a- sec- sec 6 6, is equal and thus ; ; = log a + log sec called a subsidiary 6. angle and is found from the equation log tan 6 = log b - log a. Tlius ani/ expression which can be put into the form of the of two squares can be readili/ adapted to Jogarithrnii' irori: http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight sum 215/467 , 8/12/2019 . Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 178 To 195. adapt formula the c^ =-- a^ 4- [CHAP. _ ft2 cos .2ai C to logarithmic computation. From the identities CO C= cos^ - - sin^ — cos 2 we have = (a,2 + J2) (^ ( ( 2 cos2 C _+ C and C + sin^-, =cos2— 1 2 C\ j sin2 = (a2 + 62 _ 2ab) cos2 1 + (^2 / îal 1|. ;,2 | 2 (' - - cos^ sin'- C\ j + 2ah) sin^ -^ C C = (a - 6)2 cos^ -^ + (a + 6)2 sin^ — -(«-«'-'f{i+(:4i)'-'f}. Take a subsidiaiy angle such that 6, a + 6, . , tan d = a then ^-2 = (a - 6)2 — {a - C cos2 -^ (1 6)2 co^;2 wliere is • sec2 c .'. . C — — may 6) cos C — sec^ C 6) + log cos - -j- ; log sec (f, deterinined from the equation log tan 6 = log (a -f 6) 196. ff) ; = (o- log c = log (a . b tan 2 — + tan^ C J- When two sides solve the triangle - log (a log If 2::= /-CCS2G a = 3, c + log tan — by finding the value of the third side = 1, i? = 9-9515452, 7' 48 find h = -4030862, /. tan first as in Art. 189. first = 53° -3010300, log 2-5298 33' 54 6) and the included angle are given, wo instead of determining the angles E.rample. - 2fi http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight ; given diff. for 1 33' 54 = 172, = 9-0989700. 216/467 . . 8/12/2019 Elementary Trigonometry by Hall and Knight SOLUTION OF TRIANGLES WITH LOGARITHMS. XVI. We have h- = c- + a- - = 2ca cos + C-) {a- I cos- — B — + sm- -k)-'^ ac = {a- c)2 cos^ = (a - (•)- = (« - c)- COS- -„ (1 whei'e tan 6i <1 + + sin^ ( I COS''— - Bin ' tan- 1 — tan- 0) = log 2 + log — — I = ^^-t^tan4 a - c 2 =2 log tan ^ :. + c)^ (a -I cos-— 179 •(1). tan 26° 33' /54 tan 26° 33' 54 = •3010300 + 1-6989700 = 0; whence tan = 1, and = 45°. 71 From J, (1), z= {a =2 - c) cos ~ sec 45° cos sec ' :2./2cos26°33'54 ; .-. log /) -3010300 = log 2 + ^ log 2 •1505150 log 2: 54 + log .-. log log?; cos 26° 33' log cos 26° 33' 54 = -4030902 = 40 diff. diff. for 1 is 172 40 .•. Thus the 197. -9515452 •4080902 2-5298= 40308 62 But 1 proportional increase = --^ third side is „„ ^^^• 2^529823. The formula c- = a- + 6- — '2ah cos to logarithmic computation in two other the identities cos 10 — Jo— C=2 cos^ C ^- 1 and cos http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight C may also be adapted ways by making use C= 1-2 sin^ of C ^- 217/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. ISO We shall take the [CHAP. of these cases, leaving the other as an fii'st exercise. = c^ = - 2ab cos 4- (,2 cos2 ^ - 1 + 52 _ 2ab (22cos2f-l) b'i = (a + by- — 4ah then e2 = (a + 6)2 ( .-. log c .-. To determine C ,7,2 cos^ — - cos^ e) = {a + b)'^ 1 c = (a + 6) sin ^; = log (a + 6) + log sin 2\/a^ • log cos ^ . Since 2 and less thus is a 1 = log 'i/ab is 2 + - (log a + r o <? cos 2. angle is U a = 8, b — 7, c — 9, positive lind all 5' = 9-6502809, L tan 36° 41' = 9-8721 123, C + 6) + log cos — and XVI. less — than is positive unit^^, and f. ; given log 2, diff. for 60 = 3390, diff. for 60 = 2637. log 3, difference between the angles at the base of n tri- and the sides opposite these angles are 175 and 337 the angles L tan log (a find the angles L tan 24° is 24 , ; never greater than a + b and cos than unity, cos^ an acute angle. The 2 - log b) EXAMPLES. 1. ^. we have the equation 6 ^ cos 6 = . sin2 ^ 1 2 ; given log = 93274745, 2, : log 3, L cot 56° http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 6' 27 = 9-8272293. 218/467 8/12/2019 Elementary Trigonometry by Hall and Knight SOLUTIOX OP TRIANGLES WITH LOGARITHMS. XVI.] One 3. log 2, log two-sevenths find the greater of the two acute angles given of the sides of a right-angled triangle of the hypotenuse is ; : Z sin 7, 181 14° 11 = 9-455921, ' Zsin 14° 12' = 9-456031. Find the greatest side when two of the angles are 78 and 71° 24' and the side joining them is 2183 given 14' 4. ; = -6260733, I>=103, log 4-2274 log 2-183 = -3390537. 22' 30° 14' 9-7037486. 78° sin = 9-9907766, sin = 2 ft. If 6 5. 6 in., c = 2 ft., A = 22° and then shew that the (liven log log 2, L cot sin L = 22° 20'= 1 1° 10' a If = 9, 6 L tan 29° = -43766, = -250015, log cot 29° 21' 7. Z) sin 49° 27' 34 L 10-70465, 9-57977, = 1-56234, log 56234 = 4-75, given side the other angles very approximately is <i 20', find 1 ; foot, 3, If a 6. = L L = 12, J. (7=58° 42' find 9-88079, 6 , find log cot 29° 22' = 30°j = 9-75041. = 22' 26 the A and B ; = -249715. values of c, haWng given log log 12 = 1-07918, 9= -95424, log 171 = 2-23301, log 368 = 2-56635, 30° 11° 48' 39 Xsin Zsin Xsin Z sin =9-69897, = 931 108, = 9-82391, = 9-97774. 41° 48' 39 108° 1 1' 21 sides of a triangle are 9 and 3, and the difference of find the angles ; having given opposite to them is 90° the angles ^ The 8. : log 2, L tan 26° 33' = 96986847, tan 26° 34' = 9-6990006. L sides of a triangle are 1404 and 960 respectively, and an angle opposite to one of them is 32° 15' find the angle contained by the two sides ; having given log 2, log 3, 9. Two : L 10. log 13 sin 21° 23' Uh : tan given L = 1-1139434, = 9-5621316, c=li : \{B- (7) log 1-1= cos 24° 37' 12 \0 and Zcosec 32° L -041393, cot 25', | 9-8923236. use the formula to find Xtan = 9-958607, /.tan 8° 28' = 10-2727724, sin 51° 18'= ^ = 35° = tan2 1 15' B 12° 18' and C ; 36 = 9338891, 30 = 10-495800, L cot 17° 42' 56-5 = 9-173582. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 219/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 182 [CHAP. = 1071, a = 873, find B; given log 8-73 = -941014, log 1-071 = -029789, X sin 50' = 9-884254, X sin 70° =9-972986, Xsin 70° 1' = 9-973032. If 11. ^ = 50°, 6 = 3, (7=36° 52' 12 , find c without determining A&ndiB; given log 2 = -30103, log 3 = -47712, X sin 18° 26' 6 = 9-5, log 40249 = 4-60476, Z cot 18° 26' 6 = 10-47712. If 12. a = 6, 6 {In the following Examples the necessai'y Loyarithma must he token from the Tables.) = 840, 13. Given a=1000, 14. Solve the triangle in which a = 525, 6 = 650, 15. Find the least angle when the sides are proportional to 4, 5, and 16. 6 c=1258, find B. c = 77*7. 6. If i? = 90°, ^(7=57-321, ^^ = 2858, find .1 and C. Find the hypotenuse of a right-angled triangle in which the smallest angle is 18° 37' 29 and the side opposite to it is 284 feet. 17. 18. them is The sides of a triangle are 9 60° find the other angles. : and and the angle between 7 How long must a ladder be so that when inclined to the at an angle of 72° 15' it may just reach a window 42-37 ground 19. from the ground feet a = 31 -95, ? = 21-96, C=35°, J and 20. If 21. Find B, 22. Find the greatest angle of the triangle whose sides arc C, 6 a when 6 find = 25-12, c = 13-83, .•1 B. = 47° 15'. 1837-2, 2385-6, 2173-84. 23. and When a = 21-352, 6 = 45-6843, c=37-2134, find J, B, C. 24. If 6 = 647-324, c = 850-273, ^ = 103° 12' 54 , find the remaining parts. 25. If 6 = 23-2783, A =37° 57', 7^=43° 13', find the remaining sides. 26. c Find a and 6 when A' = 72° 43' 25 , ('=47° 12' 17 , = 2484-3. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 220/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight SOLUTION OF TRIANGLES WITH LOGARITHMS. XVI.] ^6^=150, .4=31° If Ji)'=4517, 27. 30', 183 find the romaiiiiug parts. Find 28. .1, B, and b a = 324-68, 29. Given a =321 -7, 30. If 6= 1325, when c = 421-73, 6'= 35° c = 435-6, = 36° c=1665, .1 5 = 52° 19', 17' 12 . 18' 27 , find ('. solve the obtuse-angled triangle to which the data belong. If 31. ff = 3795, i? = 73° ('=42° 15' 15 , 18' 30 , find the other sides. Find the angles of the two triangles which have 6=17, c=12, and C=43° 12' 12 . 32. a triangle are 27402 ft. and -7401 ft. find the base and respectively, and contain an angle 59° 27' 5 altitude of the triangle. Two 33. sides of : difference between the angles at the base of a 17° 48' and the sides subtending these angles are The 34. triangle is 105-25 ft. and 76-75 ft.: the angle included by the given find sides. From 35. the following data = 43° .1 = 43° .1 = 43° .1 (1) (2) (3) : 15', .45 = 36-5, BC =20, 15', yl5 = 36-5, BC=30, 15', .15 = 36-5, 5r=4.5, point out which solution is impossible and wliich ambiguous. Find the third side for the triangle the solution of which is neither impossible nor ambiguous. In any triangle prove that 36. . , tan 6 a= 17-32, If 6 c = (a - = 2V^ r- sm . a— = 13-47, 0=47° o 13', b) sec 6, where C -jr 2 find c without finding A and B. 37. If If tan rf) = ^^ , tan ^ , a = 27-3, 6=16-8, ('=45° prove that 12', find http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight c = (d - b) cos ^,aud thence - sec lind (f). c. 221/467 8/12/2019 Elementary Trigonometry by Hall and Knight KLEMKNTAKY TKIGONOMETKY. 183. [CHAl'. Solution of Triangles with Four-Figure Logarithms. 197 A To , solvn a triaiiqle * when the three sides are aioeti. [For general explanation of method see Art. 187.] Example. tan .-. a =283, A ft = 317, c = 428, A ^= si(s t - 1 s (log 197 find all the angles. 285 817 197x86 /{s-b){s-c) 2-\/ log tan From If v/j514x a) + log 86 - 231' _J28 •2)1028 514=. log 514 - log 231). 231 = .s-rt s-b 197 = 86=s-f the Tables, = 2-2945 log 514 = 2-7110 86=1-9345 log 231 = 2-3636 5W46 log 197 log 4-2290 50746 2 log tan '^ log tan 20° 42' Again, tan ) 1-1544 = 1-5772 = 1-5773; ~ - 20° 42', approx., and A = 41° 24'. /(s- c){s-a) _ s{s-h) 2 V / 86x 231 514x197' 86 = 1-9345 log 231= 2-3636 log 514 = 2-7110 log 197 = 2-2945 4-2981 5-0055 5-0055 log 2 log tan ) . 1-2926 ^=1-6463 log tan 23° 48' = 1-6445 18 diff. :. for5'= 17 ^ = 2.3° 53', J = 41° 24', Thus This solution approximately, and 7J = 47°4G', may be compared [Examples XVI. g. Nos. 1 — B - 47° 46'. C=90°50'. with that of Ex. 2, Ai-t. 188. 10 may be taken here.] Any small error in obtaining the Imlf angle is doubled when we multiply by 2. Thus a half angle obtained to the nearest minute will not necessarily give a final result to the same degree ot NoTjc. accuracy. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 222/467 8/12/2019 Elementary Trigonometry by Hall and Knight >OUK-FiyUKK \V1.] If gieutfr accuracy log tan *^ log tan 20=' 36' is retjuired = i - is log tan 20 22 ditf by 36' ^ .. i'2' follows : ^ l-ollH log tan 20=^ 36' = i-5750 = 1-5750 greater than 20 ^^'û we may proceed as i-5772 diff. .-. TABI.KS. 22 ^x or 5'*7 6' for 6' . 23 •J3)i:52(o7 115 ; '- ^ I7(t *i=^20^41'-7, To 197b. ^ and ,:(=41°23'. a triangle having given two solve sides and the included angle. [For general explanation of nietliod K.ramph: ^ tan = 681, If « A-C —-— = a-c 2 a +c = 213, B = 50° c cot B = jr- 2 .see 42', solve ^38, ,.,.o.,,, pr^T cot 2o° 21'=;,:r^- tan low 438 = 2-6415 ; ^^^ i^,„ log tan 64° 39' ; By - ( = 90'^ - 2 .1 2 9660 (• ''- = 2-9657 = 64° K ^ 39'. = 109° 40', C=19°38'. and by subtraction Again, ^3345 =45°1'. ^ addition ^ •0003 '-i±^' Also 39- log 924 ..^r.., ogtan'—^— = -0003; .1 Jo j '^-^' .. ^,,^_, 64^ 39' 924 924 logtan-2- = log^38-log924 + the triangle. 438 A-C .-. Art. 189.] sin ~ sine log 243 = 2-3856 243 sin 50° 42' i>' | sin 19 ^38' log sin 50° 42' ' = 1-8887 | .-. log 6= log 243 + log sin 2-2743 50° 42' ^ -log sin 19° .-. log 6 whence Thus & .1 This solution [Ex.vMi'LK« 38'; ' log sin 19° 38' = 1-5263 ^-7480 = 2-7480; , = 559-8. = 109° may XVT. 40', C' = 19°38', be coiupai'ed g. Nos. 11 7^ = 559-8. witli that 20 may http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight Ixi on p. 172. taken here] 223/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELKMENTARY TRIGONOMETRY. 183, To 197„, [chap, a triangle having given two anglen and a solve side. [For general explanation of method see Art. 191.] Example. If h = 1000, A = 45-, C = G8'' 18', find B = 180° - 45° - 68° 18' = 66° 42' the least side = log a — log 1000 rt = h sin - log sin 66° 1000 sin 45 .1 log 1000 i 42' * sin 66° 42' Bini? log sin 45° +- the least side. log sin 45 I 2-8495 = 2-8864; log sin 66° 42' whence (( = To 2^8864 g. 21—30 may Xo.s. a triangle ichen, one of them are given. opposite to = 1-9631 769 8. [Examples XVI. 197d. =a = 1-8495 two solve bo taken here.] and sides the angle [For general explanation of method see Art. 192.] Examjile. If b b ,, = (13, sm B~- sm . , ( ' ~ c c = 86, C = 29° 23', 63 -_ 3b . sm solve the triangle. 7= -8451 29° 23' = 1-6908 log , C log sin •5359 sin 29° 23' og sin .-. 1 = A' =59° log 10'. B^ = 59° Thus we may say .-. 10', B.^ = 180°- 59° 10' -29° 23' A., = 180° - 120° 50' - 29° 23' siippleiiientaiy to the = 91° 27', = 29° 36 sinjll^7' 36sjn^8° 33^ = 1-8654. 47'. log 36= log sin 88° 33' C ~ ~sln29° 23' ~ _ ~ 1-5563 =1-9999 1-5562 log sin 29° 23' siir2li^°'23' .-. Jl = 120° 50' .li _ csinî _ sin -6021 1-9338 1-9338 Also since c^b, there is another vahio uf above, viz. 120° 50'. [See Art. 148.] * 4= =1-6908 1-8654 (f, = 73-35, from tlu; http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight Tables 224/467 8/12/2019 Elementary Trigonometry by Hall and Knight FOUH-FKiUHK TABLES. XVI.] 1 log csinJ;., ' '*-^*' _ 1-2524 a., finally log sin 10', or 120° 50', A = 91° 27', or a = 73-35, [Examples XVI. g. Nos. 29° 47', or 36-44. 31—37 may be taken EXAMPLES. XVI. (Given the three = 11-7, 5 1-5616 =86-44, from the Tables. I = 25-3, = 1-5 = 59° - 1-6908 29° 23' 28' sin 29° 1-5616. .-. Thus = 1-6961 8 6 sin 29° 4 7' ~ = 30=1-5563 log sin 29° 47' sine' B3j^ 6' = 9-0, here.] g. sides.) A nsmg sm -. . find .1, 1. If 2. If « 3. are Find the smallest angle of a triangle whose sides 11-24, r^ = 68-75, B ^ h = 93-25, c = 63, find B, tisnig cos . 13-65, 9-03, using the sine formula. = 15, 6 = 22, find all the angles. c=9, 4. If a 5. Find all the angles, given a = 31 -54, b = 46-5, c 6. Find all the angles, given a = 33-4, = 71 '6, c = 60-24. 6 largest angle of a triangle whose = 03-4. sides are 14-75, Find the 6-84, 19-37, using the cosine formula. 7. 8. angle, If the sides are 21-5, 13-7, 29-5, find the smallest using the sine formula. 9. 10. If <? 6 : : c=5 : 7 : 8, find all the angles. 1-3 If the sides of a triangle are as : 1*4 : 1-5, find all the angles. {Given 11. If 6 12. If 13. If a 11. tico sides and the included migle.) = 32-8, c= 15-0, A = 107° 26', a = 96-7, = 012, b = 135-4, c= -576, 6' -= 123° http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight B 42', find ^ = 60° 30', K. E. T. find find A and C. A and and B. ('. 1-i 225/467 8/12/2019 Elementary Trigonometry by Hall and Knight KLRMF.XT.VRY TRKiONdMETH V. 183„ K 14. sides of a triangle arc 27-3 aiul 16-8, and they angle of 45° 7', lind the remaining angles and side. Two contain an 15. Given 16. If 6 17. If 18. with it [('H.\l>. .1 = 37°, b = S2-9, c = 25-l, = 27-0, c = 33-48, .4 solve the triangle. =60°, find the other angles. 13a = 296, and C=82° A and find 14', B. and makes If one side of a triangle is double another an angle of 52° 47', find the remaining angles. 19. If 876 20. If = and 131r, = 17-6, 6 A = 18° 16', find = 24-03, C= 121° 38', and a {Given two angles 21. If ^ = 40°, (7= 70°, 6 = 100, find 22. If 5 = 42°, e=107°, a = 85-2, 23. If A =49° 24. If .1 25. If 11', 5=21° 15', c = 65° 27', ^=71° 35', J = 60°, B = 79° 20', c B and solve the triangle. side.) a. find b. = 5-23, find a. = 873, find 6 = 60, C. c. find a. Tlie base of a triangle is 3-57 inches, and the angles adjacent to it are 51° 51' and 87° 43', find the remaining sides. 26. 27. Given 5=65° 47', f=52°39', a = 125-7, find the greatest side. 28. Solve the triangle when A = 72° 19', 5 = 83° 17', < = 92-93. In a triangle the side adjacent to two angles of 49° 30' 29. and 70° 30' is 4-^ inches find the other sides. ; 30. If B C=4 : (Given : 7, tivo C=7A, and sides a = 73, 6 = 62, 6 = 89-36, and angle ^ = 82° 31. If 32. If 6 = 41-62, c = 63-4.5, /? 33. If « = 17-28, 5 = 5° 34. W<i - 94-2, 6=141 6^=23-97, -3, 14', find a and c. not inchuJed.) find 5. = 27° 15', find C. 13', find A and c. A =40\ snlvo tbc tiianglc http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 226/467 8/12/2019 Elementary Trigonometry by Hall and Knight XVI T' OUR-FlCilIlI-: J J =20' 35. If 36. If i triangle to = 1325, c = 1665, ^/ which the data belong. Find A, B, and h when 37. S3p 1 = ll;j, .solve the triangle. j5 = 52°19', .solve the obtuse-angled &= 137, 41', TABI.KS. rt = 324-7, c=421-7, C=.35°. (Miscellaneous.) If 39. r4iven 40. r a = 31-95, 38. = 12, = 21-96, = 1000, h ^=35°, = MO, = 1258, B. find B, using sin ^- Solve the triangle in wliich a = .525, /> = 650, o /* . = 17, = 777. Find the least angle when the sides are proportional to 42. and 6, using the sine formula. 43. Find B, 44. If parts. C, and a, when ?> = 25-12, J7? = 4517, .4C=:150, J =31° = 321-7, Given a 46. In any triangle prove that ( . , tail « = 17-32, ?< = 13-47, c c = 13-83, A =47° 30', = 435-f), A =36° 45. If A and find Find the angles of the two triangles whicli liave and (7=43° 12'. 41. 4, 5, re /'^ 18', find the remaining find C. = (a — 6)sec C = 2\/^ j- sni — a-b 2 15'. 6, where . . f=47°12', find c without fiiuUng .4 and B. 47. If If tan a = 27-3, ch Z) — a— , b = 10-8, tan — 2 , prove that (7=4.5° 12', find c — (a — b) cos — .sec rb. 2 (^, and thence find <;. 14—2 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 227/467 8/12/2019 Elementary Trigonometry by Hall and Knight CHAPTER XVII. HEIGHTS AND DISTANCES. Some easy cases of heights and distances depending only on the solution of right-angled triangles have been already The problems in the present chapter dealt with in Chap. VI. 198. more general character, and require are of a some geometrical for their solution a ready use of trigonometrical skill as well as formulae. Measurements To find 199. on a. the height avd in one plane. distance of an inaccessible object horizontal plane. A Let obsei-ver, the be CPthe position of the Pdraw object; from PC perpendicular to the horizontal plane; then it is required to find PC a.n&AC. At A observe the angle of elevaMeasure a base line AB tion PAC. ^ in a direct line from A towards the observe the angle of elevation object, and at B LPAC=a, Let PBG. AB = a. lPBC=^, PC=PBsm^. From hPBC, From aPAB, „ PC= a sin a sin cosec A C= PC cot a = a cos a sin . Also Each thus if AB sin PAB _ a sin a ~ sin (^ - a) sin APB • /3 . (/3 /3 of the above expressions is ' - a). cosec (/3 - a). adapted to logarithmic work PC=.v, we have log A'= log a + log sin « -|- log sin /:i http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight + log cosec O- a). 228/467 8/12/2019 Elementary Trigonometry by Hall and Knight MEASUREMENTS ONE PLANE. IX 185 Unless the contrary is stated, it ^vill be supposed that the observer's height is disregarded, and that the angles of elevation are measured from the ground. NoTK. Example I. A person walking along a straight road observes that at two consecutive milestones the angles of elevation of a hill find the height of the hill. in front of him are 30° and 75° : In the adjoining figure, PBC =75°, AB simile; /JP5 = 75°-30° = 45°. /.PAC=30°, I Let x be the height in yards ; then .T^P5sin75°; but „„ PB = —— ^BsinP^B T^=rf. — — sm APB -. —— 1760 sin 30'^ TF^— -. sm ; 45 1760 sin 30° sin 75° ~ sin 45° = 1760x|xV2x^ = 440(^3 + 1). we If is take J3 = 1-732 and reduce to feet, we find that the height 3606-24 ft. EXAMPLES. / XVII. a. above the sea-level the angles of depression of two boats in the same vertical plane as 1. From the top of a cliflF 30° 200 ft. : the observer are 45° and find their distance apart. A person observes the elevation of a mountain top to be 15°, and after walking a mile directly towards it on level ground the elevation is 75° find the height of the mountain in 2. : feet. a ship at sea the angle subtended by two forts The ship sails 4 miles towards A and the angle and B is 30°. then 48° prove that the distance of B ait the second observa3. .1 is From : tion is 6-472 miles. the top of a tower h ft. high the angles of depression of two objects on the horizontal plane and in a line passing Shew through the foot of the tower are 45°- J and 45°-hJ. that tae distance between them is 2/t tan 2 J. 4. From http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 229/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TKIGONOMETRY. 186 5. is A. [c'HAP. All observer finds that the angular elevation of a tower On advancing a feet towards the tower the elevation 45° and on advancing b feet nearer the elevation is 90° -A : is find the height of the tower. person observes that two objects A and B bear due N. and N. 30° W. respectively. On walking a mile in the direction N.W. he finds that the bearings of A and B are N.E. and due E. 6. A respectively find the distance : between A and B. A tower stands at the foot of a hill whose inclination to the horizon is 9° from a point 40 ft. up the hill the tower subfind its height. tends an angle of 54° 7. ; : a point on a level plane a tower subtends 8. a an anglean At c ft. in length at the top of the tower subtends and a flagstaflf shew that the height of the tower is angle ^ : c sin of j3 cos {a + 0). The upper three-fourths of a ship's mast subtends a point on the deck an angle whose tangent is 6 find the tangent the angle subtended by the whole mast at the same point. Example at a cosec II. ; Let C be the point of observation, and let AP being the lower fourth of it. be the mast, Let A\so\et ABÂa, AC -b, APB AP = a; iACB = e, lBCP = p, so that tan/3='6. so that From aPC^, tan(tf-/3)=p from A BCA, 4a tan ^ — -r i. 4(tau^-tanS) tan. = 4tan(.-^).î^-^--^; . .-. ; /. „. ^^' ^- lHJl_i(5tan^-^ - 5 + 3tan^ 3 1 On reduction, +- tan e tan'-^- 5 tan ^ tanS = whence ' l + or 4 = 0; 4. examples of this cia.ss we make use of right-angled triangles in which the horizontal base line forms one side. Note. The student sliould observe that in http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 230/467 8/12/2019 Elementary Trigonometry by Hall and Knight MEASUKEMEXTS XVII.] IN ONE PLANE. 187 BCD sunuounted by a siJUc /'/.' •^^fjiiids A. tower III. on a horizontal plane. From the extremity J of a ho;\x.oi;t;ii liiiu Example BA, it is BC found that CD = 72ft., and i)£ = and DE 36ft., find subtend equal angles. If BC=*J ft., BA. lBAC= lDAE = 0, lDAB = a, AB = xft. Let Now =9 i'C' , ft., , BD ^81 BE ^, ft., *^°(« + •• ^) = ££ = 117 ft. 117 lB=V = BD 81 =— X AB tana = --^ tan0 = ~— =-. AB T, . X tan(a + But , i- /,> ^)=: tana + tan(? 81x9 117.t2-81x9x 117 = 90x2; 27x2 = 81x9x117; .-. x2 = 81x39; .-. .-. But ^39 = 6-245 nearly; Thus AB = 56-2 ft. nearly. .-. A x = 9V39. x = 50-205 nearly. . long standing on a wall 10 ft. higli sul)tends an angle whose tangent is '5 at a point on the ground tind the tangent of the angle subtended by the wall at this 9. flagstaff 20 ft. : point. A 10. statue standing on the top of a pillar 26 feet high subtends an angle whose tangent is '125 at a point 60 feet from the foot of the pillar find the height of the statue. : A 11. tov/er BCD horizontal plane. BA it is prove tliat spire DE stands on a the extremity ^ of a horizontal line subtend equal angles. and From DE BC ^C'=9ft., CD = 280 BA = 180 ft. nearly. found that If surmounted by a ft, and DE=3bft., http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 231/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 188 [chap. On the bank of a river there is a column 192 ft. high 12. supporting a statue 24 ft. high. At a point on the opposite bank directly facing the column the statue subtends the same r.ngle as a man 6 ft. high standing at the base of the column : rind the breadth of the 13, point P n, -y /3, AP = .r, A monument ABODE stands on level ground. At a on the ground the portions Ji>, AC^ subtend angles = c, respectively. Supposing that AB=a, AC=b, and a + ^3 + y = 1 80°, shew that (a + 6 + c) x'- = ahc. AD Example IV. walking 1000 after rivei*. The altitude of a rock is observed AD to be -47°; up a slope inclined at 32° to the Find the vertical height of the rock point of observation, given sin 47°= '731. ft. towards it horizon the altitude is 77°. above the Let first P be the top of the rock, the points of observation A and B then ; in the lPAQ=\T, lBAC = %T, PDC= L PBE = IT, AB = 1000 ft. figure L Let X ft. be the height x=PA We of In sin then ; PAC=PA sin 47°. have therefore to find PA in terms AB. aPAB, zP^5 = 47° -32° = 15°; /JPB = 77°-47°=30°; Z.4-BP = 185°; .-. PA=^ ABP sin A PB AB sin 1000 sin 135° sin 30° := .-. 1000^2; x = PA sin 47° = 1000 ^/2 x -731 = 731^2. If we take ^2 = 1 -414, we find that the height is 1034 ft. nearly. a point on the horizontal plane, the elevation of After walking 500 yards towards its the top of a hill is 45°. summit up a slope inclined at an angle of 15° to the horizon the elevation is 75 find the height of the hill in feet. 14. From : http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 232/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight rHUBLKMS DEPENDENT ON GEOMETRY. XVli.] From 15. A B a station at the base of a 189 mountaiu its summit after walking one mile towards seen at an elevation of 60° the summit up a plane making an angle of 30° with the horizon find to another station (', the angle BCA is observed to be 135° is ; : the height of the mountain in The 16. * feet. summit elevation of the of a hill from a station A After walking c feet towards the summit up a slope inclined shew that the at an angle ^ to the horizon the elevation is y — a) feet. height of the hill is c sin a sin (y )3) cosec (y i.s a. : From 17. of Ben Nevis towards the and a point 60°; is A an observer finds that the elevation he then walks 800 ft. on a level plane and then 800 foot, finds the elevation to be 75° A Nevis above 4478 many In 200. is further up a slope of 30° shew that the height of Ben ft. : approximately. ft. of the problems which follow, the solution depends upon the knowledge of some geometrical proposition. From a tower stands on a horizontal plane. and at a horizontal distance of 48 ft. ft. the 14 plane above mound from the tower an observer notices a loophole, and finds that the portions of the tower above and below the loophole subtend equal If the height of the loophole is 30 ft., find the height of the angles. Example I. A. tower. AB Let observation, vertical We and be the tower, C L the CE horizontal. the point of loophole. Draw CD Let AB = x. have GD=U,AI) = EC = i%,BE=^x-U. From iÂDC, ^C='=(14)2+(48)2=2500; .-. AC=m. From aC£B, C£2 = (a;_l4)2 + (48)2 = a;2- 28a; + 2500. Now , , hence by lBCL= lACL; BC BL „ ^ Euc. vi. 3, —^ = -ry ; Jx^ - 28j + 250J _x-oO 50 By On squaring, 9 (x - - 28x ~ + 2500) = 25 (x^ reduction, we obtain 16a;-- 1248.(;=0; Thus the tower is 78 ft. 30 - 60.r whence + 1)00) .i-^78. high. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 233/467 ' 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TUIUONOMEXKY. 190 EXAMPLES. At one 1. XVII. fcHAP. b. on the other side of the road, at a side of a road is a flagstaff 25 ft. high tixerl the top of a wall 15 ft. high. On point on the ground directly opposite, the flagstaff and wall subfind the width of the road. tend equal angles : A 2. To an statue a feet high stands on a column 'da feet high. on a level with the top of the statue, the column and statue subtend equal angles find the distance of the observer from the top of the statue. observêr : a feet high placed oti the top of a tower h feet observer /; feet angle an the tower to high subtends the same as high standing on the horizontal plane at a distance d feet from A 3. flagstaff the foot of the tower (a Example : shew that -h)d^ = {a-\- b) A II. b^ - 2b'^h flagstaff is fixed An upon a horizontal plane. -{a-b) h-. on the top of a wall standing observer finds the that angles subtended at a point on this plane by the wall and the flagstati are a /S. He then walks a distance c directly towards the wall and finds that the flagstaff again subtends an angle ^. Find the heights of the wall and flagstaff. and ED Let A and B DC be the wall, the flagstaff the points of observation. CAD=p= L CBD, so that D are concyclic. lABD = suppt. of Z ^ CE Then z four points C, A, B, .-. = 90° + (a + Hence in /3), from , the aCAE. A ADB, lADB = 180° - a - {90° + (a -|- (i) \ :=90°-(2a + j3). AD = Hence in ABD sin ADB AB - sin in cos (a cos (2a + fi) + ^) aADE, I)E And c — AD a CA D, CD^ sin a = ,^ cos (2a - -„, + /3) ADv,mCAD _ ADwa^ ~ cos (a + ~ sin ^ CD /3) http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight csiii/i cos ( 2ft -(- (i) 234/467 8/12/2019 Elementary Trigonometry by Hall and Knight xvn.] PROBLEMS DEPENDENT ON UEOMETKY. AB ^^ vaan walking towards a tower ou which a III. is fixed observes that when he is at a point E, distant Example BC flagstaff I- from the tower, the ft. lBEC=a, c tau ( .5 Since AE subtends flagstaff prove that the heights - J 191 E and ) 2c tan a ft. of the its greatest tower and angle. If flagstaff are respectively. the point in the horizontal is which BC subtends a maximum it can easily be proved that AE touches the circle passing round the tri- line at angle, angle CBE. [See Hall and Stevens' Geometry, p. 315.] of this circle lies in the D through E. Draw DF perpendicular to BC, then DF bisects BC and The line vertical centre also lGDB. By Euc. m. 20, :. CB = 2CF = 2DF tau .: Again, L lCDB = 2lCEB = '2a; lCDF= iBDF=a. AEB— L —- ECB Z. a = 2c tan a. in alternate segment EDB at centre 1 AB — c tan AEB — c tan 1) A tower standing on a cliff subtends an angle ^ at each of two stations in the same horizontal line passing through the hase of the clift' and at distances of a feet and h feet from the cliff. Pi-ove that the height of the tower is (a + h) tan /3 feet. 4. A column placed on a pedestal 20 feet high subtends an angle of 45° at a point on the ground, and it also subtends an find angle of 45° at a point which is 20 feet nearer the pedestal 5. : the height of the column. 6. A of two places ^4 of the flagstaff If AB = a, on a tower subtends the same angle at each The elevations of the top and B on the ground. as seen from A and B are a and /3 respectively. flagstaff shew that the length of the a sin {a + (i- DO flagstaff is ) cosec (a - http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight /:i). 235/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 192 A [cUAP. At a distance of 60 feet stands on a pedestal. from the base of the pedestal the pillar subtends its greatest angle 30°: shew that the length of the pillar is 40^^/3 feet, and that the pedestal also subtends 30° at the point of observation. 7. ijillar A person walking along a canal observes that two objects are in the same line which is inclined at an angle a to the canal. He walks a distance c further and observes that the objects shew that their distance apart is subtend their greatest angle ^ 8. : 2c sin a sin A 3/ (cos a + cos /3). stands on a horizontal plane. Shew that the distances from the base at which the flagstaff subtends the same angle and that at which it subtends the 9. tower with a iiagstaft' greatest possible angle are in geometrical progression. two stations A and B subtends equal prove that angles at two other stations C and B The 10. line joining : AB sin CBD=CD sin ADB. Two 11. I straight hnes DAE= L ABC, DBE=a, and DEC meet at C. If i EBC=y, z EAB = ^, ^5sin;8sin(a+i3) BC= ^—. . shew that sm Two ^. (7-/3) o / sm (a+^ + y) , \ , • P and Q subtend an angle of 30° at A. Lengths of 20 feet and 10 feet are measured from A at right angles to ^P and AQ respectively to points R and S at each find the length of PQ. of which PQ subtends angles of 30° 12. objects : 13. A ship sailing N.E. is in a line with two beacons which are 5 miles apart, and of which one is due N. of the other. In 3 minutes and also in 21 minutes the beacons are found to Prove that the ship is sailing subtend a right angle at the ship. at the rate of 10 miles an hour, and that the beacons subtend their greatest angle at the ship at the end of 3^/7 minutes. A A stands on the top of a tower. man walking along a straight road towards the tower observes that after the angle of elevation of the top of the flagstaff is y3 14. flagstaff ; walking a distance a flagstaff subtends its the flagstaff further along the road he notices that the maximum angle a shew that the height of ; is 2a sin a sin cos /3 + sin (i (« -l:i)' http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 236/467 8/12/2019 Elementary Trigonometry by Hall and Knight MKASTTKICMRNTS TX XVTT.] MORE THAN ONK TLANK. Measurements in more than one In Art. 199 the base line 201 towards the object. AB ]93 plane, was measured directli/ as If this is not possible we may proceed follows. From A measure in a base line AB any convenient direction in the At A horizontal plane. two angles PAB and observe the PAC ; and at B PBA. Let LPAB=â, lPAC=I3, observe the angle LPBA=y, AB = a, Frovu PC=:»: A PAC, a;=PA sin j3. From A PAB, AB sin PBA „ PA=, ^xnAPB /. To 202. sheui .r=a how sin jSsin a sin sin {a ycosec (a to find the distance y + y) ' + y). between two inaccessible objects. Let P and Q be the objects. Take any two convenient stations A and B in the same horizontal plane, and measure the distance between them. At A observe the angles PAQ and QAB. Also if AP, AQ, AB are not in the same plane, measure the angle PAB. At B observe the and ABQ. In angles A PAB, we know lPAB, lPBA, so that In ABP AP may be found. A QAB, we know L QAB, L QBA, so that In l^PAQ, we A Q may AB and and AB be found. know AP, AQ, and lPAQ; 30 that PQ may be found. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 237/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRI(}ONOMRTRY. 194 Example is 18°. The angular elevation of a tower CD at a place A is 30°, and at a place B due West of A the elevation 1. due South of it AB — a, If [f'HAP. shew that the height of the tower is v/2 W. + 2V5 .-<N CD = x. Let From From But / the right-angled triangle DC A, AC =xeotSO°. the right-angled triangle DGB, BC=:x BAC is cot 18°. a right angle, .-. :. .•. BC-^-AG^ = a?\ .r2(cot218°-cot2 30°) X- (cosec- 18° - cosec- 30°) H(^0' .-. = a-; =a ; 4^=, .r2^(^/5+l)'-2-4}=a2; .-. x2(2 + 2^5)=:a2, which gives the height required. Example a road on it 2. A hill of inclination 1 in 5 faces South. Shew that which takes a N.E. direction has an inclination 1 in 7. AD running East and West be the ridge of the hill, and let be a vertical plane through AD. Let C be a point at the foot of the hill, and ABC a section made by a vertical plane running North and South. Draw CG in a N.E. direction in the horizontal plane and let it meet BF in G draw parallel to BA then if Let ABFD ; Cll is joined it GH ; will represent the direction of the road. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 238/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 239/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 196 [CHAP. A person due S. of a lighthouse observes that his shadow cast by the light at the top is 24 feet long. On walking 100 yards due E. he finds his shadow to be 30 feet long. Supix)sing him to be 6 feet high, find the height of the light from the ground. 5. The angles of elevation of a balloon from two stations a mile apart and from a point halfway between them are observed to be 60°, 30°, and 45° respectively. Prove that the height of the balloon is 440^6 yards. [If AD is a viedian of the triangle ABC, then2AD^ + 2BD- = AB^ + AC-.] 6. 7. At each end of a mountain elevation is (j). 0, and at the middle point of the base the Prove that the height of the mountain is is a sin 8. the Two of a base of length 2a, the angular elevation sin <^ Vcosec vertical poles, same angle (cf) + 6) cosec (0 — 6). whose heights are p, and h, subtend a at a point in the line joining their feet. If they subtend angles /3 and y at any point in the horizontal plane at which the line joining their feet subtends a right angle, prove that , {a + 6)2 cot2 a = a^ cot2 /3 + &2 cot2 y. From the top of a hill a person finds that the angles of depression of three consecutive milestones on a straight level I'oad are a, /3, y. Shew that the height of the hill is 9. 5280^2 10. / \/cot2 a - 2 cot2 ^ + cot2y Two chimneys AB and CD are feet. of equal height. A person standing between them in the line AC joining their bases obAfter serves the elevation of the one nearer to him to be 60°. walking 80 feet in a direction at right angles to A C he observes find their height and distance their elevations to be 45° and 30° : apart. Two who are 500 yards apart observe the bearing and angular elevation of a balloon at the same instant. One 60° and the bearing S.W., the other finds finds the elevation 11. persons the elevation 45° and the bearing W. Find the height of the balloon. 12. The horizon at an angle an angle /3 and is inclined to the straight railway upon it is inclined at side of a hill faces a. A to the horizon : if due S. the bearing of the railway he x degrees E. of N., shew that cosa.'=cotatau/i. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 240/467 8/12/2019 Elementary Trigonometry by Hall and Knight PROHLKM.S RKQUIHIXC FOUR-FIGURE TABLES. XVTT.] EXAMPLES. XVII. 197 d. [In the follovnng examples the logarithms are to be taken from. Four-Figxire Tables^ A mail in a balloon observes that two churches which he knows to be one mile apart subtend an angle of 11° 24' when he is exactly over the middle point between them find the height 1. : of the balloon i n miles. There are three points A, 2. a level piece of ground. elevation of 5° 30' from />', The angular the shore is 12° 32', ^ is,26° 34' a straight A vertical pole erected at and 10° 45' from B. line C lias If AB on an is and the distance BC. altitude of a lighthouse seen from a point on and from a point 500 find its height : in A 100 yards, find the height of the pole 3. C above the feet nearer the altitude sea-level. From a boat the angles of elevation of the highest and points of a flagstaflf 30 ft. high on the edge of a cliff are lowest 46° 14' and 44° 8' find the height and distance of the cliff. 4. : 5. the top of a hill the angles of depression of two From milestones on level ground, and in the same vertical successive plane as the observer, are 5° and 10°. Find the height of the hill 7 / in feet and the distance of the nearer milestone in miles. An observer whose eye is 15 feet above the roadway finds that the angle of elevation of the top of a telegraph post is 17° 19', and that the angle of depression of the foot of the 6. post is 8° 36' find the heiglit of the post : and its distance from the observer. Two straight railroads are inclined at an angle of 20° 16'. the same instant two engines start from the point of inter7. 1 At one along each line one travels at the rate of 20 miles an hour at what rate must the other travel so that after 3 hours the distance between them shall be 30 miles ? section, ; : An observer finds that from the doorstep of his house the elevation of the top of a spire is 5a, and that from the roof above the doorstep it is 4a. If h be the height of the roof above the 8. doorstep, prove that the height of the spire above the doorstep and the horizontal distance of the spire from tlie house are respectively /; If cosec a cos 4a sin 5a A = 39 feet, and a = and 7°19', h cosec a cos 4a cos calculate the 5a. height and the distance. H. K. K. T. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight ;|5 241/467 8/12/2019 Elementary Trigonometry by Hall and Knight CHAPTER XYIII. PROPERTIES OF TRIANGLES AND POLYOONS. To find 203, Let ABC. By the area of a tnangle. A denote the area of the triangle Draw AD perpendicular to BC. a triangle is half the area of a rectangle on the same base and of the same altitude. Euc. I. 41, the area of A=^ (base x altitude) ^^BC.AD = \BC.AB&mB = -<?« sin B. may be proved Similarly', it that A=-a6sinC, and A = ^6csinJ,. These three expressions the area are comprised in the for single statement A=Again, of two {prodîct 1 A=- . 6c sin e sides) x {sine A A=bc sin — / {s -b){8- c) V be of included angle). cos V A — Is (s - a) ^~b^ = '\/sis — a) (s — b){s — c), which gives the area in terms of the sides. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 242/467 8/12/2019 Elementary Trigonometry by Hall and Knight 199 AREA OF A TRIANGLE. 1 , A = ôc Again, 2 ' a^ sin _ 1 A —, . sin . sin 2 — —A aainC asinB — , A —. r- smA . -. . sin 5 sin C 2 sin -4 a^ sin B sin C 2 sin (5+ (7)' which gives the area in terms of one side and the functions of the adjacent angles. Many Note. writers use the symbol but to avoid confusion between symbol A is Example of lengths The 1. sides x S and preferable. of a for the area of a triangle, .ST triangle manuscript work the in are 17, perpendiculars from the angles the 25, 28: find the upon the opposite sides. From ^=9 the formula (base x altitude), dividing it evident that the three perpendiculars are found by by the three sides in turn. 2A is Now A = Js (s -a)(s- b) (s - c) = Js5 = 5x7x6 = 210. Thus the perpendiculars Example 2. and the length are -^ , -— 2o 17 x 18 x 10 x 7 ^^ , 28 , or —=17 , ô , 15, Two angles of a triangular field are 22-5° and 45°, of the side opposite to the latter is one furlong: find the area. Let From A = 221°, -8 the formula the area in sq. yds. = 45°, then A= 6 = 220 yds., and b~ sin A sin 2 sin C7=112i°. C £ 220 X 220 x sin 22|° x sin 112^° 2 sin 45° 220 X 220 X sin 22|° x cos 22^° 2 x 2 sin 22i° cos 22A° = 110x110. 110x110 1Expressed ni acres, M the area= -^.v,,;: 4840 T? —= „, 2i. If)- http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 243/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRTGONOMKTRY. 200 the radius To find 204. circumscribing a circle of the [chap. triangle. Let S be the centre of the circle circumscribing the triangle ABC, and R its radius. L BSC by SD, which Bisect also bisect BC at Now by L BSC Euc. will right angles. iil. 20, at centre twice = = 2.4; lBAC and ' = BD=BS Bin BSD = Jism A It = 2 sin sin a Example. The first A side C = 2RsmB, sin = 2R sin A, Shew b . sin = âh sin sin B that 2R- sin = - 2R A r^ = .-^=2*. = Thus ; A ^ 2R . siu B sin J3 c sin . = 2RsiaG. (7 sin = A. C (7 = A. 205. From the result the of following important theorem last article we deduce the : If a chord of length 1 subtend an angle 6 at the circumference of a circle whose radius is R, then 1 = 2R sin 6. 206. For shortness, the be called the Circuni- circle, radius the Circum-radius. circle circuiuscribing its a triangle may centre the Circum-centre, and its The circum-radius may be expressed in a form not involving the angles, for a ^ 2 sin A ^ abc 2hc sin http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight .1 _ abc ~ 4A ' 244/467 8/12/2019 Elementary Trigonometry by Hall and Knight KAOIUS OF THK INSCRIBED CIRCLE. XVIII.] Tojmd 207. a triangle. the centre of the circle inscribed in the triangle then IB, IE, IF are the points of contact and D, E, Let ABC, the radius of the circle inscribed in 201 / be F ; perpendicular to the sides. A Now A = sum of the areas of the triangles BIC, CIA, = 9«'' + = sr 9 or + ^cr = -(ti AIB + h + c)r ; whence To express 208. one side and the radius of the inscnbed circle in terms of '^ the functions of the half-angles. In the figure of the previous article, we know from Euc. iv. 4 that / is the point of intersection of the lines bisecting the v r; that angles, LIBD = \, so LICD=^\. BD = rcot^, .-. CB = rcot^. (If fix cot--t-cot. .•. . r sin = a; B B+C C sin ~ — — =asin ^ . . . a sin •• j B — 'cos ; C sm . A 2 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 245/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 202 [chap. A circle which touches one Definition. triangle and the other two sides produced is said escribed circle of the triangle. 209. Thus the We to of a be an triangle three escribed circles, one touching ABC has a second touching GA, and BC, BA produced; a third touching AB, and CVl, CB produced. BC, and AB, produced side ; shall AC assume that the student is familiar with tlie construction of the escribed circles. [See Hall and Stevens' Geonietry, For shortness, we the In-circle, radius ; its centre the In-centre^ and its radius the centres their 195.] shall call the circle inscribed in a triangle and similarly the escribed Ex-circles, p. circles Ex-centres, may be and their the In- called the the radii Ex-radii. To find 210. the radius of an escribed circle of a triangle. Let /j be the centre of the circle touching the side BC and the two sides and AC produced. AB of contact be the points Let D^, E^, F^ then the lines to these points are perpen- joining I^ ; dicular to the sides. Let rj be the radius ; then A = area ABC = area ABI^C - area BI^C = area BIÂ +area CIÂ - area BI^C 1,1 1 = -c. + 9^''r = _(c-|-5-a)ri = (s-a)ri; A s Similarly, if to the angles B r,^, and r^ — a' be the radii of the escribed circles opposite C respectively, http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 246/467 , 8/12/2019 Elementary Trigonometry by Hall and Knight RADII OF THE ESCRIBED CIRCLES. XA'III.] 203 , 211. To find tlic 1>-lv- radii of the escribed circles in terms of one side athd the functions of the half-angles. In the figure of the last article, I^ is the point of intersection ; of the lines bisecting the angles .-. BD, = r, cot (^90° CBi = ?-i cot Uo ••• n (^tan B - - and f C externally ) =î tan ~ ) so that | = r^ tan | ^ http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 247/467 , 8/12/2019 Elementary Trigonometry by Hall and Knight KLEMENTARY TRIGONOMETRY. 204 Example The Shew 1. first side — that —= — + a -^-i b s s a b \s J A s (s b s(s- -b) Ac Ac (s-c) )\ = _ + r.^ - a) (s b) Ac(s-f) s{s-a)(s-h) If J _ A(2s-a-i) A s(s - a) 2. . r„ - ^] +l( ^ ^ ^) (^ - ^l a \s Example [cHAP. - c (s + r.^ -a){s- s [s r, b) (s - c) c) prove that the triangle is right- angled. By transpositiou .A :. iR sm -^ B cos — C cos = iH . • . sin - r= i\ — ,^ - cos A/ — A . 4iv sin i — cos sin B — B — ^ i:, + r.^ B . . C — sm — sin G cos -^ ,^ + 4tR cos B C — — sin — sm A — cos B — . sin C — . cos 2 „ 2 2 2 B B A / C G\ sm - cos - + cos - 0.x.-., sm = cos —.«xx.-ou.,--rv.u.2 . . ( .A— sin cos 1 B+C —~— = . whence ^ = 4-5°, and ^ — .,^4 sm- cos A — sin . B+C — z^ ; .,A —cos- — = 90°. Many important relations connecting a triangle and circles may be established by elementary geometry. 213. its With the notation of ])rcvi()us articles, circle from the same point arc equal, http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight since tangents to a 248/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 249/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELKMENTARY TRIGONOMETRY. 206 EXAMPLES. J I. Two sides of a triangle are 300 included angle >/ 2. XVIII. is 150° ft. [cHAP. a. and 120 and the ft., find the area. ; Find the area of the triangle whose sides are 171, 204, 195. > Find the sine of the greatest angle of a triangle whose sides are 70, 147, and 119. 3. 4. of the If the sides of a triangle are 39, 40, 25, find the lengths perpendiculars three from the angular points on the opposite sides. ' are One side of a triangle is 30 22|° and 112|°, find the area. 5. 6. ft. and the adjacent angles Find the area of a parallelogram two of whose adjacent and 32 ft., and include an angle of 30°. sides are 42 7. Xj ^ angles 1y The area is 150° : rhombus 648 sq. yds. and one of the find the length of each side. of a 8. In a triangle 9. Find if r^, r^, r^ a= 13, is &= 14, = 15, c find r and R. whose in the case of a triangle sides are 17, 10, 21. If the area of a triangle is 96, and the radii of the escribed circles are 8, 12, 24, find the sides. 10. Prove the following formulae II. s/rr^r.^r^Â. 13. rr^cot 15. r^r^r^ 17. 72r(sin^+sin^ + sin 18. r-^;r^ 20. r,+r.^ : — = A. a) ta,n— s{s 14. 4Rrs = abc. 16. r + rr^ = ah. (7) cot— cot — = r^. = A. 19. con— \/bc{s-b){s-c) = A. 21. i^)\-r){;r.^ Q 2- = A. C H = rs^. = ccot — 12. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight + r.j^ = aK 250/467 . . 8/12/2019 Elementary Trigonometry by Hall and Knight XVIll.] RKLATIOXS OF A TRIANGLE AND 2 22 —= cot — = r„ r, cot 23. - H 25. ri+r2+r3-r = 4R. 27. 62 sin 26^+(;2sin 28. 4^ 22. r., cot -7:=r cot 222 — cot — cot — ~ h 26. r..r.j r a^-¥ .1 sec s'-. + i\ + r.,-r._^=-'i.licoiiC. - i)' —^ — 2Rciim{A-B). —7a— = —n2— •~sinAsinB sin (A - B) a'-b^ 31. If the perpendiculars from A, B, =o^ p.,, ^ ^- Cio the opposite sides pg respectively, prove that i + l=l; (I)'i + Pi Pi Ps Pi ^ {ri-r){r^-r){i-^-r) rj\r \r i + i-i.i (2) Prove the following identities 32. + rir., 2^ = 4 A. -30. Avepi, + r.,r, = 29. » -= C^ . == 24. — = (a + 6) cos 207 ITS CIRCLES. P'l = ARr\ rj ~ rj\r rh'^' 4A(cotJ+cot5 + cot(7) = a2 + 62 + c2. „_ b 35. ^1 c + —a a— + ^2 ^3 '^?. : 34. —c Pi ^ h = 0. + sin 2B + sin 2C) = 32A^ 36. aWc^ 37. acosA + bcosB + ccosC=4RsinAsinBsinC. 38. acotA+bcotB+ccotC = 2(R+r). 39. (b-\-c) (sin 2.4 A tan — + (c + a) tan Ji C - + {a + h) tan - = AR (cos A + cos B + cos C). A + sin B + sin C) = 272 40. r (sin 41. cos2-+cos^-+cos-2 = 2 + sin ^ sin B sin C 27i. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 251/467 . 8/12/2019 . Elementary Trigonometry by Hall and Knight ELKMENTARY TRIGONOMETRY. 208 [CHAI . Inscribed and circumscribed Polygons. To find 214. 11 the perimeter sides inscribed in &. poli/goa of circle. a Let r be the radius of the AB and area of a regular circle, and side of the polygon. Join OA, OB, and draw OD bisecting then is bisected at right AB lAOB; angles in D. And L A 0B= - (four right angles) Jin n Perimeter of polygon = iiAB = 2nAD = = 2nr sin Area of polygon = n (area of 1 = ^ wr^', sm . 2 2nOA sin A 01) - n triangle ^1 OB) — 27r n 2h find the perimeter and area of a regidar polygon of 215. n sides circumscribed about a given circle. Let r be the radius of the a side of the AB the circle at £>. OD bisects AB bisects polygon. Let Join OA, OB, sA, circle, AB OD right angles, ; and touch then and also lAOB. Perimeter of polygon = nAB = 2nA D = 2n0D tan A OD = 2nr tan - . n Area of polygon = « (area of triangle AOB) -^nOD.AD /*/•'-' tan • http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 252/467 8/12/2019 Elementary Trigonometry by Hall and Knight TNRORIBEP AND C'TRCnMSCRIBBIP POLYGONS. XVTTT.] 209 no need to burden the memory with the formulfe of the last two articles, as in any particular instance There 216. is they are very readily obtained. Example 1. The side of a regular radius of the circumscribed circle. Let r be the required radius. adjoining figure we have AB = dodecagon 2 is find ft., the In the iAOB = ~. -2, AB = 2AD-2rsin'' Thus the radius N/2(x/:-i ^3-1 sin 15° J6 + J2 is + l). feet Example 2. A regular pentagon and a regular decagon have the same perimeter, prove that their areas are as 2 to ^/5. Let .4J5 polygon, circle, OD Then be one of the n sides of a regular the centre of the circumscribed perpendicular to AB. if AB = a, area of polygon = nA D . :=nAD. — na- OD AD cot , -T- cot ir — . 4 n Denote the perimeter of the pentagon and Then each side of the pendecagon by 10c. tagon is 2c, Each and its area is 5c- cot side of the decagon is Area of pentagon Area of — . o _ decagon ~ c, and its area - c^cot is 2 cot 36° _ 2 cos 36° sin 18° cot 18° ~ sin 36° cos 18° _ 2cos36° _ 2(V5 + 1) i ~l + cos36°~ + ^ 2(^5 1) ^ 5 + v'5 . _ . 2 cos 36° 2 cos- 18 / \ — sja ^ + l 4 2 x/5' http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 253/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 210 217. To find the area of a [CHAr. circle. Let r be the radius of the circle, and. let a regular polygon of n sides be described about it. Then from the adjoining figure, we have area of polygon = n (area of triangle A OB) ndAB.OD = \0D.7iAB - X perimeter of polygon. increasing the number of sides without limit, the area and the perimeter of the polygon may be made to differ as little as we please from the area and the circumference of the circle. By Hence area of a circle T =- x circumference = -x27rr 218. To find the area [Art, of the sector of a r>9.] circle. Let 6 be the circular measure of the angle of the sector then by Geometry, area of sector _ 6 ' area of circle 2tt area of sector i i'TV EXAMPLES. XVIII. b. ~1 22 T)i thi.f V 1. Kvercise taki he 7r= ^ . Find d the area of a regular decagon inscribed whose radius is 3 feet ; given sin 36° = in a cin^le •r)88. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 254/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight INSCRIBED AND CIRCUMSCRIBED POLYGONS. XVTTI.] 211 Find the perimeter and area of a regular quindecagoti given described about a circle whose diameter is 3 yards 2. ; = -213. tan 12° that the areas of the inscribed and circumscribed a regular hexagon are in the ratio of 3 to 4. Shew 3. circles of *^ 4. Find the area of a circle inscribed whose area is 250 sq. ft.; given cot 36° = 1 5. circle t^ pentagon 376. Find the perimeter of a regular octagon inscribed whose area is 1386 sq. inches given sin 22° 30' = 382. Find the perimeter of a regular pentagon described about 6. is 616 sq. ft.; given tan 36° = 727. Find the diameter of the circle circumscribing a tiuindecagon, whose inscribed circle has an area of 2464 7. given sec 12° V in a ; a circle whose area 1-^ in a regular regulnisq. ft. = r022. Find the area of a regular dodecagon in a circle about a regular pentagon 50 sq. ft. in area given cosec 72° = 1 '0515. 8. ; A 9. regular pentagon and a regular decagon have the area, prove that the ratio of their perimeters is tjh Two 10. regular polygons of same perimeter ; shew that the 2 cos - n : n sides and 2n : same >/2. side.s have the ratio of their areas is 1 + cos - n . 2a be the side of a regular polygon of n sides, /? and r the radii of the circumscribed and inscribed circles, prove If 11. that ^ + r=acot-— 2n Prove that the square of the side of a regular pentagon 12. inscribed in a circle is equal to the sides of a regular hexagon sum of the squares of the and decagon inscribed in the same circle. 13. With reference to a given circle, Ai and B^ are the areas of the inscribed and circumscribed regular polygons of n sides, A.2 and are corresponding quantities for regular polygons of prove that B.^ 2n sides : (1) A^ is a geometric mean between A^ and (2) B^ is a harmonic mean between http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight /I, and i/, ; /i,. 255/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 256/467 . . 8/12/2019 Elementary Trigonometry by Hall and Knight THE KX-CENTKAL TRIANOLE. XVIII.] *220. AVith To find and the sides angles of the ex-central triangle. of the last article, tlie Hgiire I 213 BI^C= L = / /. BIJ+ BCI+ L CIJ CBI M----IThus the angles are 90°-^, 90°-f, 90°-f. Again, the points B, z .•. .•. G are concyclic: 71/2/3= supplement of .•. /<,/n '' .sides ^ = triangle. are similar ; = acosec — = 4/icos-— 2 2 are 4ytcos — iKcos — , . Z A JL To I^BC= iI^BC = cosec-; «^H^^ -^j 4/icos^, *221. ^ I^BC the triangles /1/2/3, •••*=7rC Thus the I.^, /j, the find area and circum-radias of the ex-ccntral The area = - (product of two = ^x iR cos ^ OD9 cos ^ — = oB? cos 2 B — cos sin ( 90° - „ ) C — 2 -l/icos- // . x (sine of included angle) 4^ cos — X 2 The circum-radius = — 2 X sides) ^J^^ r sm /,/,/, * ^ ^ = , „ . 2 sui /„^, ( A\ = 2A'. 90 - - j H. K. K. T. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight l(j 257/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 214 To jmd *222. With the the distances between the in-centrc e.v-centres. IB 7^, ICIy are right angles ; diameter of the circum-circle of the triangle JJCI^ /7i is the BG ^î-r; sin a A 57, C sin — 4Rii'm — = 4/2 Thus the and figure of Art. 219, the z s .•. [chap. cos- . distances are 4R sin ^ , , -ili sin — ^ z Ji The Pedal Triangle. *223. Let G, H, K be the feet of the 2>eqjendiculars from the angular points on the opposite sides of the triangle is ABC then ; OHK Pedal triangle of called the ABC. The three perpendiculars A 6', which BH, CK meet in a point is called the triangle Orthocentre of the ABC. *224. To find the sides and angles of the jyedal triamjle. In the figure of the last article, the points K, 0, U, concyclic .-. Also the points .-. are LOGK=LOBK=%(f-A. iT, 0, G, C are concyclic ; LOGH=LOCn=^0'-A; .-. Thus the angles B lKGR=180'-2A. of the pedal triangle are 180 - 2J, 180° -25, http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 180-26'. 258/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 259/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 216 *227. angle Ill we have seen that Art. 219, [chap. ABC ifs the pedal tri- ex -central triangle of its Certain theorems depend- /jZj/j. more ing on this connection are evident from the adjoining iigure, in which the fact that ABC is the brought more prominently into view. For instance, the circum-circle of the triangle ABC is the nine points pedal triangle of IiIÎz is and circle of the triangle Iil^^^, passes through the middle points of 11^, IJ,. 11.^, 11^ and of //g, To find *228. IÎ^, the distance between the in-centre and circum- centre. Let S be the circum-centre and /the in-centre. Produce to meet the circum-circle in join CH and CI. IE Draw perpendicular Produce AC. H ; to meet the X, and join CL. ^^S circvimference in A to Then lHIC=ÎAC+lICA ~ A 2 c ^ 2 ' lHCI=lICB-^lBCU J^-\-L 2^2 BAH ' .-. .-. Alh lHCI=lH1C; EJ=IIC=2Rsmi AI^ IE come— -^r cosoc -^ ; AI.lII-îRr. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 260/467 8/12/2019 Elementary Trigonometry by Hall and Knight DISTANCES PROM CIRCUM-CENTRE. XVIll.] 2Vi Produce SI to meet the oircumferenee in J/ an(i X. Since 3fJN, AIH are AI. chords of the IH=MI. IN={R + Sr) {R-Sl) 2Rr=B?-SI^; .-. that circle, SP = R^-2Rr. is, To find *229. the distance of an. ex-centre from the circum- centre. Let S be the circum-centre, and then A I pi-oduced I the in-centre passes through the ex-centre I^. ; Let AI^ meet the circum-circle in H; join CI, BI, CE, BH, CI^, BI^. Draw IÊ^ perpendicular to AC. HS to meet the circumand join CL. L, Produce ference in The angles IBI^ and ICI^ are hence the circle on 11^ as diameter passes through B and C. right angles ; The chords circum-circle at BH CH of the equal angles and subtend A, and are therefore equal. But from the last article, HC= HI; .-. hence H is HB=^HC^HI; the centre of the circle round IBl^C. HI^ = HC=2R Now SIi^ — ii^ = square Hin A of tangent from 7^ = IiH.IÂ = 2/fc sm — . Ty cosec — = 2Rr^. .-. SIy^ = R:^ + -2R,\. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 261/467 . 8/12/2019 . Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 218 To find *230, the distance [chap, of the orthocentre from t]ie circum- centre. With the usual SO' ' we have notation, = SA + ^ 02 - %SA .AO cos SA 0. 2 lôw AS=R; AO = AHGoseGC = c cos A C = 2R sin C cos A cosec C cosec = 2E cos A ; lSAO=lSAC- lOAG .-. = (90°-^)-(90°-C) = C-B. SO^ = E'- + 4E^ cos'- A -4R'- con A cos {C-B) = R^- Am COS A {cos (5 + C) + cos (C- B)) = m- 8i?2 cos A COS 5 COS C. The student may apply a two articles. ÊXAMPLES. 1. Shew tliat sm — Shew that the distances ajigular points A, B, Care C - cos 2 3. 2 'iRs A 7-- cos 2 V — - (2) ; A , . f. 4/fsni , 2 cosec is /, — from the A B 2 2 — cos— is e^pi.il tn — cosec — cosec Sliew that r 5. . sm A . Prove that the area of the ex-central triangle (1) 4. „ 4/t , — sin the c. of the ex-centre 2. B — XVIII. the distance of the in-centre from 4/1 4/r cos to establish method similar results of the last . //i . //, . 7/3=47? . TA Shew that the perimeter and . rn . IC. iii-i'iidius of tlic ])ed;d triangle are respectively All sin A sin li sin 6' and 2/? cos http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight A cos H cos C. 262/467 8/12/2019 Elementary Trigonometry by Hall and Knight THE EX-CENTRAL AND PEDAL TRIANGLES. XVIII.] If 6. h, 5', k denote the sides of the pedal triangle, prove that m ^' 9 '^ a2 ^ , k^ a^ + b^ + c^ 62^-^2 2a6c a- ' h- c~ Prove that the ex-radii of the pedal triangle 7. •iH cos A sin 219 B sin C, 2R sin A cos 5 sin ai'e 2R sin J (^, sin J5 cos C. Prove that any formula which connects the sides and 8. angles of a triangle holds by a, b, c (1) and A, B, and A B, , we replace bcosB, «cos^4, C by by «, i, c (2) if ]80 -2J, a cosec C c by ^ 90° , - 6', 180°-2/?, b cosec ^ cos — 90° , c , - f cosec - , Prove that the radius of the circum-circle than the diameter of the in-circle. 9. shew that the triangle 10. If R==2r, 11. Prove that *S72 4- SI^^ is iH0°--2(': 90° - , ^'. never less is equilateral. + SI.;^ + SI./ = 1 27?2, Prove that 12. (1) a.Ar^ + b.BP + c.CP = abc; (2) a If 13. . AI/ - OHK b . BI^' - c . 01/ = abc. be the pedal triangle, and C) the orthocentre, prove that OS OK AG^ BH^ CK~ qa_ ^' OH OG+a cot A ^ OH+b cot B OG ^ ' ' ^ GHK OK r;/r+c cot 0~ be the pedal triangle, shew that the circum-radii of the triangles AHK, BKO, CGH 14. If http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight tlie is ' sum of equal to 263/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 220 [chap. If AiB^Ci is the ex-central triangle of ABC, and A^B^fio, 15. the ex-central triangle of A]B^(\, and A^B^L\ the ex-central triangle of ^2^2^ 2> ^^^ ^^ *^ î^d ^^ angles of the triangle J„^„6'„, and prove that when n is indefinitely increased the triangle • l)ecomes equilateral. Prove that 16. = 9i22_a2_52_c2. OP = 2r2 - m^ cos A cos BcosC; OIi^ = 2r^^ - AR^ cos A cos 5 cos C. 0>S2 (1) (2) (3) h denote the distances of the circum-centre of the pedal triangle from the angular points of the original ti-iangle, 17. If y, (/, shew that 4 (/^ +5^2 + h^) = 1 li?2 + 8^2 cos A cos B cos 0. Quadrilaterals. *231. To prove that the area of a quadrilateral is equal to - {^produet of the diagonals) x {sine of included angle). A C, BD interL DPA = a, and Let the diagonals sect at P, let *S' and let denote the area of the quadrilateral. ADA C= AAPD + ACPD = ^DP.APHina + 1 DP. PC HiniTT- a) ^^DP{AP + PC) sin a DP.ACmna. AABC=^BP. AC sin a. Similarly -. S=^{DP+BP) AC sin a D B.AC sin a. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 264/467 8/12/2019 Elementary Trigonometry by Hall and Knight QUADRILATERALS. XVIII.] To find *232. and a quadrilateral in terms of ABCD lengths of be the quadrilateral, and its sides, «, Ji, d be the e, By equating the two values of Blf^ found from the triangles D, BCD, we have .-. Also C a^ + d^-b^-c^- = 2adcosA-2bccosC sum *S'= 16/S^ . Let (2) be sin (1). C and add to the square of (2). (1) + (a2 + c^2 _ j2 _ c2)2 = iaW + ib^c^ - Sabcd cos (A + C). J + C= 2a ; then cos {A + C) = cos mS^ = A{ad+bcf - .-. ; 45= 2ac/ sin .4+ 26c sin r .-. Square 2bc cos of areas of triangles 5.4/), Bf'D =~ ad sin A +- • let its area. >S' a^+d^-âd cos A=^b'^+c^ - . the sides mm of tiro opposite angles. the Let BA the area of 221 But the first {a^ 2a =2 cos^ a - + d^- ¥ - c^)'^ 1 - \&abcd cos^ a. two terms on the right = (2arf + 26c + 0.2 + d^-b^- c-') {2ad +-2bc- a^ -d'^ + b- + c^-) = {{a + d)^-{b- c)2} {{b + c)2 - (a - df} = {a d+b-c){a = (2(r+- 2c) {2a- - c){b + d-b-+ 2b) (2a- 2d) + -+ {2a- c-a a-d){b c + 2a), where o d) + + 6 + c4-o?=2o-, = 16 (o- - a) (o- - 6) (o- - c) (o- - d). Thus where S'^ = {a- - a) denotes the of opposite angles. 2a- {a- ~b) sum of - c){a- - d) - abed cos^ a, the sides, 2a the sum of either {a- In the case of a cyclic quadrilateral, tliata=-90°; hence *233. S=^J{ar-a) {a-—b) This formula may {a- — c) {a + t'=180'', so — d). be obtained directly as http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight .4 pair in the last article 265/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 222 [CHAP. by making use of the condition ^-f-C=180° during the course of In this case cos C= - co.s J., and sin f'=sin A, so that the work. the expressions (1) and (2) become a2 a'2 - 62 - + and To *234. 6^2 + /^'?<c? = 2 (ad: 4*S'= 2 whence by eliminating 1 c-2 A we (_,,2 be) cos {ad + be) sin A A , ; obtain + ^2 _ ^,2 Âe diagonals _ c2)2 = 4 (ac^ + 6c)2. and the circum-radius of a (Ufcllr qnadrilateral. If ABCD is a cyclic quadrilateral, 2 {ad+ be) we have just proveil that A = a^ + d^-b^- c\ cos Now BD^ = a^ + d^- 2ad cos A a<^(a2+c?2-62-c2) = a2, + c<^ ,, V <- aa-\-oc bc{a^+d^) . ^— i + ad(b^ + c^) ad+be . _ {ab + ed) {ac + bd) ad+bc Similarly, we may prove that .po-^ :^ +bc){ac+bd) ab + cd AC.BJ) = ac + bd, 'I'lms AC _ad+bc BD ab + rd and The circle triangle A BD ., . passing round the quadi-i lateral circumscribes the ; hen(;e BD .. the circum-raauis= ^-. 2 ~ sm A (ad +be) ~ 2 (arf + be) BD _{ad+bc)BD si n A'' AS = -T-^ '\/{ab + ed) (ac + bd) (ad + be). http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 266/467 8/12/2019 Elementary Trigonometry by Hall and Knight QUADRILATERALS. XVIII.] A Example. inscribed in Â . tan2 5- 2 = it quadrilateral and another ABCD circle 223 such that one circumscribed about is circle it shew that ; be -J ad . a circle can be inscribed in a quadrilateral, the sum pair of the opposite sides is equal to that of the other pair; If Since tlie quadrilateral A= —2 (ad— j-T— + jr-^-j [Art. 23.1]^ . 6c) a-d = h-c, so that a^ -2adJrd^ = b^-2bc :. one of is cyclic, cos But can he rt2 + c^-; + d2-&2_c2=2(ad-fcc); cos^ = „A tan- — =1 2 1 ad- he ad-\-he^ A be -= —J ad + cos A - cos XVIII. EXAMPLES. d. If a oircle can be inscribed in a quadrilateral, shew that its radius is S/a- where >S is the area and 2a- the sum of the sides 1. of the quadrilateral. 2. If tlie sides of a cyclic quadrilateral be 3, 4, 4, 3, shew that a circle can be inscribed in it, and find the radii of inscribed and circumscribed circles. 3. If the sides of a cyclic quadrilateral be 1, 2, 4, 3, tlie shew 5 that the cosine of the angle between the two greatest sides is _ , I and that the radius 4. The sides of the inscribed circle is 98 nearly. of a cyclic quadrilateral- are 60, 25, shew that two of the angles are right angles, and diagonals and the area. 5. is 9 : 6. The sides of a quadrilateral are 4, 5, 8, 9, 39 .52, find : the and one diagonal find the area. If a circle can be inscribed in a cyclic quadrilateral, that the area of of the circle tlie quadrilateral is i\J ahcd, and tliat shew the radius is 2 \/ ahcd I {a + h-\-e + d). http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 267/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 224 7. area is [CHAP. If the sides of a quadrilateral are given, a maximum when shew that the the quadrilateral can he inscribed in a circle. 8. If the sides of a quadrilateral arc 23, 29, 37, 41 inches, prove that the maximum area is 7 sq. ft. 9. If A BCD a cyclic quadrilateral, prove that is ^ 2~{a-e){<T-d)' If /, ff denote the diagonals of a quadrilateral angle l)etween them, prove that 10. 2fff 11. lateral, cos /3= (a2 + c2) ~ /i the + d^, If ^ is the angle between the diagonals of any quadriprove that the area is ^{(a2 + 6'2)~(62 12. (62 and + c;2)}tan/3. Prove that the area of a quadrilateral in which a can be inscribed circle is vabcd A+C I 2 If a circle can be inscribed in 13. diagonals are /and g, prove that a quadrilateral whose 4S-^=fY-{ac-bdy. 14, If (jiiadrilateral, 15. the angle prove that is i3 + (1) [a^ (2) cos ^= {3) tan Iff, 2 (a2 diagonals of a the ^ = {ad + be) sin c}clic A; + c2)~(52 + c;2) 2(ac+bd) - ^^^^^^ (^ _ c) y are the diagonals of a ,S'= 16. sin bcT) between i V4/292 _ («2 or (^ _ fe) (^ _ d) quadrilateral, + c2 - 6-^ • shew that - d^K In a cyclie quadrilateral, prove that the produt't of .segments of a diagonal abed ttio is {lie + hd)i{ab-\-cd) (od http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight + bc). 268/467 8/12/2019 Elementary Trigonometry by Hall and Knight following exercise consists of miscellaneous (jucs- The 235. tioiis 225 MISCELLANEOUS EXAMPLES ON TRIANGLES. XVXU.] involving the properties of triangles. XVIII. EXAMPLES. 1. e. If the sides of a triangle are 242, 1212, 1450 yards, .shew that the area is 6 acres. and the of the sides of a triangle is 200 yards find the area. jacent angles are 22-5° and 67-5° 2. One ad- : = 29-2 = 2?-3 shew that 3a = 4ft. 3. If )\ 4. If a, 5. Find the area of c are in a. p., b, y z 6. If shew that sin A 6'-, c- «-, , shew that r^, r.^, r^ are a triangle whose sides are sin : z z + X-, X + X 3S y y -, - C= sin {A-B): are in A. in h. v. V + -• z sin (^ - C), P. Prove that C_ a sin A^-h sin B-Vc sin 4 cos „ 8. 9. / — a2 . \sin (r., A — cos S C ^4 C ^ B \ A c' + -^-„ + -^-r, sm-sin-sin- = A. z 2 2 sm B sm CJ + rs) (rj . r^) (r^ + r.^) r,^ , tan 11. be cot 12. (- 13. The perimeter in-radius is ^ +mcot- + a6 cot ^ -+6 : + . . = 4B (r./j A + tan C= B + tan ^ ^ 10. + + c- 2s — cos — fe2 7 a^ + b^ r^>\ ' ' + r«+r, =4i?s2 + ly.^. -^ • U+^+ 7) • -=--+- + of a right-angled triangle is 70, and the find the sides. If/, g, h are the pc^-pendiculars from the circum-cciitre on the sides, prove that 14. a b c _ ahc http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 269/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 226 and a regular hexagon have the shew that the areas of their inscribed circles All equilateral triangle 15. {fame perimeter are as 4 to : 9. *16. Shew that the perimeter *17. Shew that the area 18. Cj , of the pedal triangle of the ex-central triangle a5c(a and [CHAP. is is equal to equal to + 6 + c)/4A. In the ambiguous case, if J, a, h are the given parts, Cg the two values of the third side, shew that the distance between the circum-centres of the two triangles — is -r^. ° be the angle between quadrilateral, shew that *19. If /3 sin^= *20. Shew *21. Shew that the sum \ . ^ the diagonals of a cyclic -j-j. that ex-central triangle *22. 2 sin is of the squares of the sides of the equal to 8i2(4^-fr). can be inscribed in and circumscribed about a and if /3 be the angle between the diagonals, .shew If circles (juadrilateral, that cos /3= (ac 23. If /, //*, ~ hd)l{ac + hd). n are the lengths of the medians of a ti-iaiiglc, prove that ( 1 A{1;^ (2) (62 (3) 16 24. + m^ -\- 7*2) = 3 (aH ^2 + c-) + (c2 - a2) to2 + (a2 _ 62) {I* + m* + n*) = i){a* + b* + c*). - c2) l-i Shew that the ,;,2 radii of the escribed _ ; circles are the roots of the equation .r' -{AR + r) aP' -\- s-x - s^r = 0. A3 be the areas of the triangles cut off by tangents to the in-circle parallel to the sides of a triangle, prove 25. If Aj, A2, that _ J^^2 Aj ( _ A3 _ A ~ ?^ (JT^yz - (sT^ Ag http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight • 270/467 8/12/2019 Elementary Trigonometry by Hall and Knight MISCELLANEOUS EXAMPLES ON TRIANGLES. XVIII.] The *26. triangle LMX 27. J, />, formed by joining the points of is shew that it is similar to the ex-central and that their areas are as r'^ to 4^-. contact of the in-circle triangle, 227 ; In the triangle PQR formed by drawing tangents at ('to the circum-circle, prove that the angles and sides are 180 180° -2,^, a and -25, -2^; 180° h c 2cosCcosJ' 2cos5cosC 2cos^cosi?' If />, J, r be the lengths of the bisectors of the angles of a triangle, prove that 28. 1 ,,, (1) - cos ^ 2 j9 ,^s pqr 5 1 + - cos + - cos - = C^ 2 5' {b a -H 1 r 6 +c AG, BH, CK are produced to the circum-cii'cle in Z, i/, N, prove that 8 A cos A cos area of triangle (1) LMN— (2) ; + c){c+a){a + b)' If the perpendiculars 29. 2 ?• 1 1 abc(a + b + c) _ 4A 1 meet B cos C AL sin A + BM&in B + CN .sin C= SR sin AsmB si n C. be the radii of the cii'cles inscribed between the in-circle and the sides containing the angles J, B, C respeccO. tively, (1) *31. ABC angle If 'i'a, )' ,, )\ shew that r„=.rtan2^- + V'-c'-o + v''V(, = r. of a triangle the pedal triangle form a trishew that the perimeter and area of A'TZ are parallel to : \/?'6'*c drawn through the angular points l.ines XIZ (2) ; the sides of respectively •2R tan *32. A A tan B tan C and R' tan A tan B tan ( '. straight line cuts three concentric circles in A, B, and passes at a distance p from their centre tlic triangle formed by the tangents at A, B, : of BC. CA . C shew that the area C is AB 2p http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 271/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIC40N0METRY. 228 [cHAP. MISCELLANEOUS EXAMPLES. + ^ + y + S=180°, shew that cos S = sin a sin cos a cos /3 + cos F. If a 1. -y /3 + siii y siii d Prove that 2. cos (15° -.4) sec 15° Shew 3. sin (15° — J)cosec 15° =4 sin J. that in a triangle cot A + sin A same vahie retains the — if cosec B cosec C any two of the angles A, B, C arc interchanged. a = 2, 4. If 5. Shew b = JS. A = 30°, solve the triangle. that (1) cotl8° = V5cot36°; (2) 16 sin 36° sin 72° sin 108° sin 144° = 5. Find the number of ciphers before the 6. digit in (•0396)»o, log 2 first significant given = -30103, log 3 An = -47712, log 11 = 1-04139. observer finds that the angle subtended by the lino joining two points A and on the horizontal plane is 30°. On 7. B walking 50 yards directly towards find his distance from 8. Prove that 9. Shew that B A the angle increases to 75° : at each observation. cos^ a + cos- /3 + cos^ y + cos^ (a +/3 + y) = 2 + 2 cos ifi+y) cos (y + a) cos (a +/a). = 2 sec 10°; 70° + tan 20° = 2 cosec 40°. (1) tan 40°+ cot 40° (2) tan Prove that 10. (1) ,- (2) 2 sin 4a -.sin lOa . SHI 27r . 47r „ +M111 -„ — + sin 2a= IHsin (jcosacos2asin-3a Gtt sni-zr- sui _ sni . =4 , . http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight n . 37r „ sni ; hn - . 272/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 273/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELBMKNTARY TRIGONOMETRY. 230 21. If (1+cosn) (l+cos/3) 22. +cosy) = (l -cosu) expression that shew (1 is equal to +sin each Tf the sum (I (1 -cosy), sin sin a of four angles is 180°, -cos/3) j8 y. shew that the sura of the products of their sines taken two together is equal to the sum of the products of their cosines taken two together. *23. In a triangle, shew that (1) 24. II^.lL.n^ = lQRh; (2) 1I{'-Î^1^^=\&RK Find the angles of a triangle whose sides are proportional to http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 274/467 8/12/2019 Elementary Trigonometry by Hall and Knight CHAPTER XIX. GENERAL VALUES AND INVERSE FUNCTIONS. Thk 236. 6=n—^, 6 equation sin^ and all =- is satisfied angles co terminal with i^y these =-, will and by have the example shews that there are an infinite niunber of angles whose sine is equal to a given quantity. Similar remarks apply to the other functions. same This sine. We proceed to shew how to express by a single foi'uuila angles which have a given sine, cosine, or tangent. all From the results proved in Chap. IX., it is easily seen that in going once through the four quadrants, there are two and only two positions of the boundary line which give angles 237. with the same Thus the arc if sin a has a given value, positions OP sine, cosine, or tangent. of the radius vector and OP' bounding the angles a and n-a. If cos a positions of [Art. 92.] has the a given value, the radius vector are OP and OP' bounding the [Art. 105.] a and 2ir - a. angles has a given value, the positions of the radius vector are OP and OP' bounding the angles If tan a a and n + a. [Art. 97.] 17—2 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 275/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGOXOMETRV. 232 To find a formida for 238. all the angles [chap. which have a given Let a be the smallest positive angle which has a gi\en sine. Draw OP and OP' bounding the angles a and it — a; then the required angles are those coterminal with OP The and OP. positive angles are 2p7r+a and where p is zero, or any 2.j)7r + {Tr — a), positive integer. The negative angles are -(rr + a) and -(27r — a), and those which may be obtained from them by the addition that is, angles denoted by of any negative multiple of 27r ; and 2qrr-{Tr+a.) where q is zero, may - {^tt be grouped as follows 2p7r + a,] .^^^ (2q~2)n + aJ a), (•(2^ \(2? ' : + l)7r-a, - 1) tt --a, be noticed that even multiples of n are folhiwed by and odd multiples of tt by - a. will it + a, - or any negative integer. These angles and 2qTr Thus all angles equi-sinal with a are included in the formula + (-l) a, any integer positive or negative. '«7r where /i This i.s zero, or is cosecant as Example The also the formula for all angles a. 1. Write down the general solution of sin least vahie of 6 the general solution Kxample which have the same 2. which is nir + ( satisfies 9 = the equation is - ; ^ . therefore 7r 1) ,, . Find the general solution of sin2^ = sin-a. This equation gives either sin <?= +sino or siu0— - sin http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight a— sin ^1), (- a) (2). 276/467 8/12/2019 Elementary Trigonometry by Hall and Knight GENERAL VALUE OK EQUI-COSINAL ANGLES. XIX.] From = n7r+(-l)«a; e-mr + (- ) {- 233 e (1), and from (2), Both values are included in the formula d To jind a formula for 239. a). = mr^ a. which have a given all the angles cosine. Let a be the smallest positive angle which has a given cosine. Draw OP and OP' bounding angles a and 2 tt - a then the the those coter; required angles and OP'. niinal with (JP The positive angles are and 2jt)7r+a whei-e are ipir is zero, ji> or + {In - a), any positive integer. The negative angles are -a aud -(27r-a), and those which may be obtained from them by the addition of any negative multiple of 'iqiT where q is zero, The angles or — 27r a and may angles denoted by — (2:7 - + '''1 a), integer. be grouped as follows f(2i' and 2^rr-a,J but is, iqw any negative ^^'' and that ; : + 2)7r-«, \{2q-2)7v + will be noticed that the multiples of it be followed by may Thus angles all +a or by - equi-cosinal a, tt are always even, a. with a are included in the formula 2?J7r where n This is zero, 1 . Find the general solution of cos ^ = - ~ least value of ^ is tt - 2«7r ± — 27r TT , o IS same a. Example The or any integer positive or negative. also the formula for all angles which have the is secant as ± a, or ; o . hence the general solutioi; . http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 277/467 8/12/2019 Elementary Trigonometry by Hall and Knight KLKMRNTARY TKIOONOMETRY. 234 To find a formula for 240. all the angles [OHAP. which ham a given tangent. Let a be the smallest positive angle which has a given tangent. Draw OP and OP' bounding the then the a and tt + a required angles are those coterangles ; ininal with The OP and OP'. positive angles are where jo is zero, or 'ipn +a any positive integer. and 2p7r + (7r The negative angles are — {it — a) and — (27r + a), — a), and those which may be obtained from them by the addition of any negative multiple of 27r that is, angles denoted by ^qiT -{it -a) and ^qn - (Stt - a), ; where q is zero, or any negative integer. The angles may be grouped + a, I (2j-2)7r + a, J 2i>7r as follows : + «, |(2y-l)7r + a, r(2^+l)7r ^^^^ whether the multiple of tt is even or odd, it is always followed by + a. Thus all angles equi-tangential with a are included in the formula and it will be noticed that 6 is This = nTT-\-a. also the formula for all the angles whieji liave the as a. same cotangent Example. Solve the equation cot id The general solution whence is 4d~mr + 30 = 7nr, 01 6 6 ; = All angles which are both 241. = cotO. 3 equi-sinal with a are included in the fonmda Snjr and equi-cosinal + a. All angles equi-cosinal with a are included in the formula But in the formula 2«7r±a so that the multiple of tt is even. «7r + (-l) a, which includes all angles equi-sinal with o, when the multiple of tt is even, a must be preceded by the + sign. ; Thus the formula is 2?i7r + a. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 278/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight 235 GRNERAL SOLUTTON OF EQUATIOXS. XTX.] In the solution of equations, the general value of the angle should always be given. 242. Solve the equation cos 9^ = cos 56 - cos Example. By (cos 9^ transposition, 2 cos 5^ cos 4^ .-. .-. .. either cos 5^ From the = 0, + cos 6) - or = 0; -cos 5^=0; cos5^(2cos4^-l) = 0; 2cos4^-l = 0. equation, first cos 5^ bd 4e and from the second, T = 2mr^- = ^2?t7r±- EXAMPLES. or , ^ or 6 , , 6= (4nil)7r ^ ± 1) = {6« z-^ XIX. ; TT . a. Find the general solution of the equations 1. 9. : sin^ = -. 4. 7. 10. 12. 14. 16. -sin 2<9- sin ^ = + sin 18. sin 19. cos ^ 20. sin 50 cos 6 21. sin 11^ sin 4(9 22. ^2 23. sin7^-x/3cos4^ = sin^. 24. l+cos(9 = 2sin2(9. 25. tan^ 26. cot2^-l=cosec^. 27. cot 28. If 2 cos fl 29. If sec = J2 4(9 + cos cos 3^ 3(9 3^ - + cos 5^ + cos 7^ 0. = 0. = sin 60 cos 2d. + sin 5(9 sin 2^ = 0. cos ^ = cos 5^. = - 1 and 2 sin 6 and tan (9 = ^/S, = - 1 , (9 + sec - tan find find http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight (9 (9 (9 =1 = 2. 0. 0. 279/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 236 In the following examples, the solution 243. the use of some particular Example 1. cosm^ = cos where k By is zero, any or transposition, may is - ( md = sin ii0. - h^ -- 1)0 zero or any ( or (m-n) e=y2k --Air. be solved through the also - -md = sin \ medium of tlie n6\ '^-me=pTr + {-\)Pnd, :. p by we obtain sin where simplified integer. (m + n)e=(2k-\-^Tr, This equation sine. For we have aoii ?»6i=:2/.-7r± .-. is artifice. Solve the equation Here [CHAP. integer + {-l)Pn]e={^^-pîr. [,n The general solution can frequently be obtained in several ways. The various forms which the result takes are merely diiferent modes of expressing the same series of angles. Note. Example 2. Solve ^/3 cos^ Multiply every term by - , + sin ^=1 then v/3 2 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 280/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight GENERAL SOLUTION OK EQUATIONS. XIX.] 237 In examples of this type, it is a common mistake to but this process is objectionable, because it square the equation introduces solutions which do not belong to the given equation. Note. ; Thus in the present instance, JS cos 9 3 cos- ^ by squaring, But the solutions - sin e =1 = (1 - sin 6) . of this equation include the solutions of - ^/3 cos ^=1- sine, as well as those of the given equation. Example^. From (oos^ e + sin + sin S)(cos 6 -%\\\d] cos d - sin or (1), From (2), + sin = eo?, ? ^1 + sine; = (1), 0=1 (2). cose + sine either From f?. cos-d- sin-e = cos# this equation, .-. .-. cos2e = cos Solve tane=-l, 1 —- - cos v' ^ 1.1 -^ sm V^ 1 .-. cos <? n cos - - sin d 4 cos • = -^ V- sm • 7 4 ; = 1 -771 v'2 (''^j) = 72^ :27(7rrt 4 / = 2n7r or 2H7r EXAMPLES. XIX. b. Find the general solution of the equations : + cosH^=0. 1. tanp^ = cot^5. 2. .sin/«^ 3. cm6-sl'is.m6=\. 4. sin 5. f,)s^ 6. sin(9+v/3co.s(9 = v/2. = ^'3(l -sin^). http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight ^-^3 cos (9=1. 281/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIOONOAfETRY. 238 Find the general solution of the equations ^- sin ^ = -—. 7. cos 9. cosec^ + cot^ = v/3. 2 sin ^ sin 3(9=1. 13. tan ^ + tan 3(9=2 tan 15. cosec ^ + sec ^ = 2 ^/2. 17. sec 19. 1 20. tan3^ + cot3(9 = 8cosec3 21. sin B 22. cosec B = s/Z cosec 0, 23. sec 4(9 -sec 2(9 + V3 tan2 ^ = ( 1 + ^3) tan Explain the equations 24. why sin 14. cos ^ 16. sec 6 - cosec 6 = 2 ^/2. 18. cos 3^ + 8 cos^ (9 .3^ =8 - sin 2<9 sin^ (9 ^. = cos 2(9. = 0. 6. 2(9 (/>. cot ^ = 3 cot 0. cot ^ = v/3 cot = J'2, sec ^, 2. 12. + 12. ^3 cos 6 = ^1^ cos = V^ sin (^, = cot ^ - cot + ^/2 = 0. 10. 2^. =2. : cos(9+siu^ 8. 11. [PHAP. <^. the same two series of angles are given by = 2%7r±^. e+4 ^=nn 25. Shew and b + i-\Y~ 4 (9-^ .3 that the formulae '2n + ^^7r±a and (^n - tt ^) +(- comprise the same angles, and illustrate hy a 1) (|-«^ figure. Inverse Circular Functions. If 244. sin^ = *-, we know that B may be any angle whose convenient to inversely by writing B = sin ~ ^ «. sine is s. It often is express statement this In this inverse notation B stands alone on one side of the equation, and may be regarded as an angle whose value is known through the medium of its sine^ tan~^V3 indicates in a concise form any one of Similarly, only whose tangent fornuila «7r + -- is ^/3. . Thus ^ = tan But all \/3 tliese and ^ the angles are given angles by the = »7r+- are e(iuivalent statements ex))ressed in dift'erent forms. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 282/467 , 8/12/2019 Elementary Trigonometry by Hall and Knight INVERSE XTX.] FUNCTIONS. Kxpressious of the form cos 245. called Cinf'iri,.\R ^.', 239 sin^'a, tan W> are Inverse Circular Functions. must be remembered that these expressions denote and that —I is not an algebraical index that is, It angles, ; sin~i .r is —X not the same as (sin x)~^ or —. . sin 246. infinite From we Art. 244, number an inverse function has an see that of values. /denote any one of the circular functions, and /~ i (.r) = J t1ie principal value of /î(.t) is the smallest numerical value of A. Thus the principal values of If sin-i(^-~V C0S-1-, are cos-ir--^j, -30°, 60°, tan-i(-l) -45°. 135°, Hence \i x be positive, the principal values X all lie between and 90°. of sinî.r, cos i.r, tan~i If X be negative, the principal values of sin~i.r and tan~ij; between and — 90°, and the principal value of cos~ ^ x between 90° and 180°. lie In numerical instances 'principal value If sin 6 247. is we usually suppose shall lies that the selected. — x^ we have cos 6 = \/l -.r-. Expressed in the inverse notation, these equations l)econie •^'^' — ^x, cos~ ^ 1 — 6 smr = V In each of these two statements, 6 has an infinite number of A'alues but, as the formulae for the genei-al values of the sine and cosine are not identical, we cannot assert that the equation ; sin~ is identically 1 This true. X = CO.S will 1 be seen 1 numerical instance. If Vl - -^ more , ^ = -, thenvl — •*' = clearlv from a x/3 ^ • Here sin~i.r may be any one of the angles 30°, and COS > v 1 - •*' 150°, 390°, 510°, ...; naay be any one of the angles 30°, 330°, 390°, 690°, http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight .... 283/467 . 8/12/2019 . . Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 240 From 248. we may the relations established in the previous chapters, deduce corresponding relations connecting the inverse functions. Thus in the identity 1-tan'^^ --„ „, 1+tan-^^' ^. cos 2^=:, let [CHAP. = a, tan so that ^ = tan~i a — ; then l-g- . 2 tan- . a = cos * ' , :, \+a- Similarly, the formula cos 36 = 4 cos'* ^ -3 cos ^ wlien expressed in the inverse notation becomes * 3 cos 1 rt = cos ~ ' (4a^ - 3a). To prove that 249. tan ~'^ a; + tan ^ i ?/ = tan ~ ~ ^ . \-xy ta,n~^x = a, so that tana=.r; Let tangly =/3, and We recpiire Now rt + i3 in so that tanj3=y. the form of an inverse tangent. tan(a+/3) tan a = 1 a th at i tan ~ 1 s, T)V ])utting + tan - tan a tan — + /3-tan-i, x + tan ' y :^- = tan - . ^ ^ y = x, we obtain 2 tan ^ ' .*; = tan 2.t' ' 1 No'iK. )3 It Ik nsoful to remember tan (tan ' .)• - ., X- that ^' = :j+ tan~' y) _ http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 284/467 . 8/12/2019 . Elementary Trigonometry by Hall and Knight INVERSE CIRCULAR FUNCTIONS. XIX.] Example Prove that 1. tan~' 5 - tan The 241 = first side tan^-^ tan~i : 1 1 + 15 J- + 3 + tan~i - = «7r + - tan tan~i 8^ 'J tan 7 = tan-i 1 72 = nw + Note. some The value -r . 4 n cannot be assijûed until we have selected of particular values for the angles tan~'.5, tan~'3, tan 7 '^^ If we choose Wig principal values, then u—0. Example Prove that 2. 12 ,4 . 1() 65 We may write this identity in the form . , 4 + Sin~' ^ Let a — sin ^ 4 - , ,12 COS-l :r^ sin : 16 so that sin a = and We cos (a bo = 16 cos 65 , ; 5 12 , so that have to express Now TTP 4 5 12 /3=:cos~'-— lo ^ tt + ;8 cosji=-. lo as an inverse cosine. + p) = cos a cos /3 - sin a sin /3 whence by reading off the values of the functions from the figures in the margin, we have / cos(a ov 3 + ^): 5 12 13 4 £ 5 13 _16 ~65' http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 285/467 , 8/12/2019 . . Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONUMKTIIY. 242 It is sometimes convenient to -work entirely [CHAV. in terms of the tangent or cotangent. Example Prove that 3. 2 cot i7 The first —-+ side^cot ^^^ 2x7 = cot ^ 24 -^ + 7 = . cot^^v 4 ;^ cot ^ T 4 24 3 7 4 cot-i A 125 H + cos^^- = cosec~^^- , 24 3 T +4 44 125 _, , = cot-ijp^=coseci— 1 EXAMPLES. XIX. Prove the following statements ly 1. sm-i- = cot ^^. 3. sec (tan ~ ^ .r) = a/i + : 2. 4. 6. tanî~ + cot-i--- = tanîg 7. cot- 8. 2tan-i-+tan = 1 tan i-- 32 = tan-i-._. 115 7 43 4 + tan-i - = tan ~ .5 +tan- 1 4 84 jg 3* ]5 4 2 3 = tan~i-. .1 i--cot-i-- = cot 5 10. taii 24 2 1 3 ^-— '^. / tan i --tan~i 3 - 2 tan~i-T = tan I 5. tan co.secî-— 1 •^ • 4 9. c. ' o + i ^j b + tan-' tan-' ,„-cot — 1 11 '3. lo http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 286/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight INVEKSK CIRCULAR FUNCTIONS. XIX.] = tan-'--. i- + sin-i 5 5 11. tail 12. 2cot-i|=tan-f. 14. sill 15. cos 16. 2tan 17. 2tan-i^+tan-iJ + 2tan-i^ = ^. - *) = (2 sill - 1 .V =2 sin li 2 tan 13. - A^.in- ^—2 ^ 1 /- = cos-i^^'. a + :c \/ a 1 4 5 / 18. sin-ia-cos-^'t^cosî (6^1- «^ + «Vl 19. sin-i-+cos~i 4 5 2 = cot-i-— 1 ;^/5 cos-i— + 2tan-i- = sin-i 21. t.an-1 w. 65 -. 5 5 + t,aii-l -w.^rns-i . \/(l i20 ,16 , + /«2)(l+«2) _il596 23. cosy|-cos-.^^ = |. 24. tan (2 tan 25. tan-ia = tan-i^ - '- x) ^2 - 1 .r tan (tan -^x + tan - ^ i-+tan-i + ab i If tan -^'^j' 2 -,- 20. 26. ||'. vTÂ'^. 2.1- 8 oo 243 + tan ~ + tan ^ , 1 ^ « + bc = tt, a;-''). +ta,u-^c. prove that y 27. If u = cot 1 \/cos n - tan ' Vcos a, prove that sin tt^tan-g. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 287/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 244 250. We shall now shew how [cHAP. to solve equations expressed in the inverse notation. Example 'Sir Solve tan~i'2.T 1. tan i3.i-=H7r + Wc have tan — ~-—mr+—-\ ' - bx- 1 2a; + 3x tan 6.1-2 -5;c- = 0, 1 4 / ^ ; .-. •4 BttN 7(7r H ( oi- (6.r Example By + = - , 1 l)(.i:-l) ; = 0; 1 --. , ...r J- or = l, Solve sin~^x + sin~i(l-.i-) = cos~ia;. 2. transposition, sin ^ (1 - x) — cos~^x - sin i.r. Let cos~ij;=:a, and sin~ix = /3; then sin~^(l -x) 1 . . But also cos a — X, sin - X = sin (a - /3) and therefore — a-/3; = sin a cos sin a ^ — .r, and therefore cos /3 - cos a sin /i. = a^/1 - x-' ^— ,>Jl- x'-; 1-x=:(1-x2)-.t2= 1-2x2; :. .-. 2x ~x = Q; x = 0, or -. whence EXAMPLES. Solve the equations 1, sin -i.r= cos 3. tan-i(.r 4. cot 5. sin 6. cos 1 X 1 .r 1 .i' - ' XIX. d. : *.«:. tan i.?-=cot â.\ 2. + l)-tan-i(.r-l) = cot-i2. + cot _ 1 - cos ~ ^ - sin ~ 1 2.r a.- = —4 g^r = sin ~ X — cos . - ^ (3.^' ^ xjZ. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 2). 288/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight IXVERSR CIRCULAR FUNCTIONS. XTX.] tan 7. - ^+ tan - 1 x-2 — 1 = ,-- x+2 245 4 3 8. 2 cot~i 2 9. tan - A- 1 + cos~i + tan - 1 — ar ^-^2 - ^ (1 ,l 10. cos - 1 11. sin 2a -, = cosec~i.r. 5 ^^ ~ - .v) =-. 2 tan il ^ 2.V , 2 tan - 1 x^. x. 1 l-fe2 -— , .,=cosî ,o. 1+0- l-x- a- 1 — ^^r,i = i^^rp r--^, + tan-i.^ 1 \/x - ^ + — ~+ 2x T- 1 - 12. cot- 1 13. Shew that we can — sin-i-^j tan .r ^:-- = ^ A in the ^^ + d^ and y are rational functions of If sin [2 cos 15. If 2 tan 16. If sin - (tt 1 - (cos ^) cos ^) If sin cot ^) (n- = tan - = cos (tt = cos (tt cot 5) = cot 1 0. c, = 0, ^ , , . , .1-2+?/- d. find x. (2 cosec ^), find 0. » shew that 3 sin-i--. 4 tan is form sin a, b, ^)}] sin ^), (tt that either cot 20 or cosec 26 If tan - {cot (2 tan 1 2^= + 18. Air express c^ 14. 17. + X^—\ =-„4-sin-i-^ aHfc^ where 2.V 1 6), and n of the form tV tan 0), is any — - is any 4 and n integer, shew integer, shew . that ^ tan ^ = 2n + — r 4 19. Find all \/4n2+ 471-15 l 1- — - ~ . -. 4 the positive integral solutions of tan ~ 1 .r + cot 1 1/ = tan H. K. E. T. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight ^ 3. 18 289/467 8/12/2019 Elementary Trigonometry by Hall and Knight KI-EMENTARY TRIGONOMETRY. 246 MISCELLANEOUS EXAMPLES. G. If the sines of the angles of a triangle are in the ratio of 1. 4:5:6, shew that the cosines are in the ratio of Solve the equations 2. 3. tan/3=2 If sin a sin 2 : 9 : 2. : 2cos3(9+sin2^-l=0; (1) 1 sec^^- 2tan2(9=2. (2) y 00860(0 + 7), P^'o^^ that cot«, cot/3, cot y are in arithmetical progression. 4. In a triangle shew that 4r 5. (r^ + Pro\'e that o sm-i- (2) (6c -— tanî - tan 1) + rg) = 2 + ca + ab) - {a- + h^ + c^). 1113 ?-2 + tan ^ - = tanî Y -- ; i 1 + sin 1— +sm 1-=-. Find the greatest angle of the triangle whose sides are given log 6 = -7781513, 185, 222, 259 6. ; Xcos39° rr 7. 8. angles 9. TP It , tan / 14' /,\ (a-\-d) = 9-8890644, =n If in a triangle a right angle. is , , tan (a SR'^ - ^y 6), drawn AliCJJ ., = a^-\-lfi-\-c'^, . = 1032. —— = slU 2^ -. sin 2a ?i- ?i - 1 +l . prove that one of the of n sides inscribed in a of the area of the circumscribed regular iwlygon with the same number of sides 10. l' prove that The area of a regular polygon circle is three-fourths for diflf. : find From n. B the straight lines 45° the direction of to to two boats are each inclined at is a straight sea-wall. the wall, and from (' the angles of inclination are 15° and 75°. If i? 6'= 400 yards, find the distance between the boats, and the distance of each from the sea-wall. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 290/467 8/12/2019 Elementary Trigonometry by Hall and Knight CHAPTER XX. FUNCTIONS OF SUBMULTIPLE ANGLES. 251. Trigonometrical ratios of 22^° or ^. From the identity o = l-cos45°, 4 sin^ 22|° = 2 - 2 cos 45° = 2 - ^2 2sin2 22^° we have 2sin22^° .-. = \/2-V2 (1). In like manner from 2 cos2 22^° 2 cos 22^° we obtain = l+cos 45°, = V'2 + ^2 (2). In each of these cases the positive sign must be taken before the radical, since 22^° is an acute angle. Again, tan 22i^° = — —sm45 -. .-, 252. We cos 45° ]^ --r— = cosec 45° - cot 45° tan22F = x/2-l. liave seen that 2 cos- 8 = V2+-s/2 ; 4cos2:^ = 2 + 2cos^; but 16 .-. 8 4cos2^ = 2 + V2T72; .-. 2cos^ = v/2 + V2+72. imilarly, and so 2 cos ^ = a/2+s/2 + \/2 + ^2 ; on. 18->2 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 291/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRTOONOMETRY. 248 Suppose that cos 253. A = - and that [f'HAP. required to find is it A sm-. . cos A 1 '2=^V _ ^ V / 2^ V'~2y \/ V 2 V 4~-2' -2 4 This case differs from those of the two previous articles in All we know of the angle A is that the datum is less precise. contained in the statement that its cosine without some further knowledge respecting is ^1 we cannot remove the ambiguity of sign in the value found for sin We now -^ . proceed to a more general discussion. Given co&A 254. equal to -, and to find sin A ^ A and cos— and. to explain the presence of the two values in each case. From the identities 2 sin- A y = 1 - cos A and , 2 cos^ A — = 1 + cos A we have A . «i«2=-V /l -cos J. 2 Thus corresponding for sin — , A fl+co&A 2 '^^Î^-V 1 , ' • to one value of cos J, tliere are two values and two values for cos — . two values may be explained as follows. If cos A is given and nothing fm'ther is stated about the angle A, all we know is that A belongs to a certain group of eqiii- The presence cosinol angles. to cos this -- group, we of these Let a be the smallest positive angle belonging then A=2nn±a. Thus in finding sin— and are really finding the vahies of sin5(2njr±a) and cos - http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight (2w7r ± a). 292/467 8/12/2019 Elementary Trigonometry by Hall and Knight FDNCTIONS OF SUBMULTIPLE ANULES. XX.] Now sin-(2?(7r±a) = .sin = sin f 249 /i7r±g) iin + cos nn- cos sin - ^ = ±sin2, for sin nn = . and cos 1 / T cos - (2nn . Again, Thus there cos A -^ jitt when are cos A = ± 1. ± a) = cos nn cos I \ two vahies is for a - sin _ + sin nn .a . ~ and sin ^ two given and nothing further is vahies for known respecting A. Geometrical Illustration. Let a be the smallest angle which has the same cosine as A then 255. tive posi- ; A = 2mr±a, and we have and cosine to find the sine A of — , that is ^ of http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 293/467 , 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMKTRY. 250 A A [CHAI'. between certain known limits, the ambiguities of sign in the formuke of Art. 254 may be removed. If COS 256. Example. A ^ sm — and . A — If cos cos and given, is 7 -^, 25' lies and A lies between 450 ^ and 540°, find A — A_ /I - cos 2~V A_ /t 2 flT _ ~ ~]\25J~ V2V fl + cosA _ f^25~ V 3 11.- ±-. /YfVZlS- 5 Now — 2 between 225° and 270°, so that sin lies ' - and cos — are both negative ^ sm — = . .•. 4 - , A 3 —= 5' A A 257. and cos , To find sin— awo^cos— m ^<jr?rts o/sin J and to explain the presence of four values in each case. We have and 2 sin / P.y addition, Viy subtraction, .-. and By A A — + cos^ u = 1> sin^ A A — + cos A\' I sin :^ ( sin ^— sin ^ ^ — + C0S — = sin ^ —— ^ ^ cos cos '^ ) ^ is | = ^ + *^ ^ '' 5 = 1 — sin A. ±\/14-sin J — = ± vl addition and subtraction, since there A — = sin A. we -si obtain sin J (1), ^ (2)- A — A and cos ^— ; and a double sign before each radical, there are four http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 294/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight FUNCTIONS OF SUBMULTIl'LE ANGLES. XX.] A ~ values for sin value of sin , and four values for cos A — SAJ corresponding to vac ^1. The presence of these values four may be explained as follows. A given and nothing else is stated about the angle A all we know is that A belongs to a certain group of equi-sinal angles. Let a be the smallest positive angle belonging to this If sin is group, then jl=n7r we ai-e + (- Thus l) a. in finding A A sin^ and cos^ really finding sin -{?i7r First suppose sin- {«7r + (-l) and a}, cos - n even and equal + (- ) «} = sin = 2m to (wtt sm miT {?i7r + - 1) ( a}. then ; + ^j cos - + cos run sm - = ±sm-, siu?«7r=0, since and cosm7r=±l. Next suppose n odd and equal to 2m + sin - {nn + - 1) a} = sin = sm ... ( is In like manner cos — known it may +9 ~ \ /TT viTT • Thus we have four values nothing further mir cos (^ - ( then ; n IT ( 1 2 — a\ n ) /TT . . + ' ^''*'^ « *'' «^ for sin — when sin A is given and respecting A. be shewn that has the four values ± cos ^ , http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight + cos ( „ — g ) 295/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOJIETRY. 252 Let a be the G-eometrical Illustration. which has the same sine tis A then 258. positive angle and we have 1 is + (- n 2»i + l, A — that , is of )- a] even and equal to 2m, becomes rnn this expression If + {-iy''a, and cosine of to find the sine -^{htt n smallest ; A=iim- If [chap. +^ . and equal to the expression becomes is odd '^'^+'|-i The angles denoted by the formula WTr + g are bounded by one of the lines OPj^ or OP2 and ; those denoted by the formula of the lines OP3 Now sin XOP^ = sin - sin XOP,= - sin XOP, = mn + i^-^) are bounded by one or OF^. ; - , sin sin ( XOP^ = „ — sin ^ XOP^ = - sin XOP., = - sin Thus the values of sin — ; ( j - ly are ±«m(|-^y ±sin^ and Similarly the values of cos ^ are + cos ^ and +cos http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight ( - - ^ j 296/467 8/12/2019 Elementary Trigonometry by Hall and Knight 253 FUNCTIONS OF SUBMULTIPLE ANGLES. XX.] If in addition to the value of sin 259. A we know that A between certain limits, the ambiguities of sign in equations and (2) of Art. 257 may be removed. Example i -- Find sin 1. and cos A — iu terms of sin A when .1 lies (1) hes between 450° and 630°. ~ In this case From the adjoining figure between — cos between 225° and 315°. lies and evident that greater than is is negative. A sm — + . .•. A sin— these limits and is it sin — cos A - = - — - cos „ :. / , ;^l + sm^, - sjl - sin A. 2 cos Exaviple order that —= - ;^/l + sin 4 + >/l - sin A. Determine the limits between which 2. 2 cos The given ^s/l- sin A, Jl + sin-= and sinA- - 2 ^ ^ - ^1 + sin 24 relation is sin From (1), we lie in Jl- sin 2 A. obtained by combining sin4 + cos4= - and - A must 4 -cos4= + see that of sin V-'- + sin24 v/l - sin A and cos 24 (1), (2). A the numerically greater is negative. From (2), we see that the cosine is the greater. Hence we have between numerically greater than sin A which cos A is and is negative. A lies to choose limits From between 2mr + ~ 4 the figure we see that and 2nir+-r. 4: http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 297/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONUMKTKV. 254 Example H. Trace the chauges of cos d - sin d in sigu and magnitude as d increases from to - sin ^ = cos 27r. ^2 — ^/2 I ( — from 6 increases TT from to j increases from 6 j creases numerically from As +1 \ - sin $ sin 7 j 1 . ... the expression 6 increases from to — to - -j- decreases numerically from - As increases increases from As is positive and decreases To find Since cos 9° from is negative and in- ^^''2. ô -r- ^2 the expression , is negative and to 0. — — 2rr, , the expression is positive and to expression is positive and the 1. the sine > sin 9° and .-. the expression to ^^2. increases 260. , -^ ô from decreases from ^/2 to and and is cosine of 9°. positive, we have sin 9 + cos 9° = + \/l + sin 18°, sin 9° — = — ^1 — sin 18°. cos 9° sin9'' + cos9°=+^l+^^5zii=+^/^3:fr75, sin 9° -cos .: and (e sin 1 to 0. As and , —^ cos cos 6 cos = ;>,/2cos s [c'HAl . 9°=- f^ l-^—^ = --^5~J5. sin 9° = i {V3+V5 - cos 9° =- [^Jz V-'i^x/s], + ^l^ + sJo-Jb]- http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 298/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight FUNCTIONS OF SUBMULXIPLK ANGLES. XX.] EXAMPLES. 1. A= — . 8ni 2. If cos A = tween 270° and 3. If cos a. between — 270^ and — 360^, prove that ^1 lies When XX. 255 r—^ /l — cos4 find , iby sin — a and cos ^ when A lies bc- and cos ~ when A lies be- 2 360°. A= — ™-r , find sin ^ tween 540° and 630°. 4. A — A ^ in terms of sin ^1 when A lies Find sin^ A and cos^ A in terms of sinÎ when — A lies in terms of sin A when A lies Find sin and cos between 270° and 450°. 5. between 225° and 315°. 6. Find sin A — and cos A — between -450° and -630°. If sin A A 24 7. J=--, 25' find • cos-- when J sin— 2 and 2 lies between 90° and 180°. 8. If sin^=-;—r, tween 270° and 9. iiiid sin ^ and cos ^ when .1 lies be- 360°. Determine the limits between which A must lie in order that (1) 2 sin ^ = V 1+ sin 2^ - Vl - sin 2A (2) 2 cos A=- \/r+sm2T+ Vl-sin2^ (3) 2 sin .4 = - V iTsin 2 J -f- \J 1 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight - ] sin 2 J 299/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 256 10. If = 240°, ^1 2 sin If not, 11. how must it = 's/l+sinA -\/l ''— be modified coi-rect -sin ( J.. ? Prove that (1) tan7|° = x/6-V3 + V2-2; (2) cotl42|'' 12. Shew that 13. Prove that 14. the following statement is [cHAP. sin 9' lies (1) 2sinir (2) tan When = v'2 + x/3-2-s/6. ir 15' between Ibd and = n/2-\/2 + n/2; = a/4 + 2v/2 - 15' from varies IS?. (x/2 + 1). to 27r trace the changes in sign and magnitude of cos 6 + sin (1) AVhen magnitude of 15. varies from to tt, 261. ^ A — and 2sin^-sin2<9 tan(9-cot(9' To find tan ^3 cos 6. trace the changes in sign tan^+cot5_ ^ sin d - (2) ; ^ whe7i tan 2sin5 + sin2^' ^ ^1 is given and to explain the the two values. presence of Denote tan Ahy t then ; — 2 t = tan ^1 = 1 .'. , ••• A of — tan- — A A Uan2^ + 2tan--«=0; ^-^^^^ The presence tan- -2±^/4 + 4<2 = -l±Vl+it2 1 -Yt these two values may be • explained as follows. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 300/467 . . 8/12/2019 Elementary Trigonometry by Hall and Knight 257 FUNOTTONS OF RnBMTTT-TTPLE ANOLKS. ^X.l be the smallest positive angle which has the given the value of tangent, then A=nn + a, and we are really finding If a tan-(w7r+a). Let (1) tan - (mr + = tan n) / 1 - tan tan («7r + a) 2m + \; a\ TT raw ( = + ^ + 1. U A = 170'. , 5 A , /tt I + , ^^ n and ., A -= ^ ^ ^ then = tan ) Thus tan - has the two values tan Example then ; + 1 j = tan ( »i7r Let n be odd and equal to {-2) 2m be even and equal to 71 prove that tan a\ • ,_, j /^ tan I ^ + is an acute angle, so that tan in the formula -~ ^ ^ ^ ^ * ^^°' tan^ ^ ^ . ^J -1- Jl + ta.n-A ^^^-^ positive. Here - X Hence must be the numerator must have the same But when ^ = 170°, tan^ is negative, and numerator therefore we must choose the sign which wil l make the sign as the denominator. negative ; , tan thus Example 2. A - 2 + tsLn-Â = -l-Jl .„„ tan Given cos A = -6, A • , find tan -^ , and explain the double answer. „A ^^'^'2 .-. Here all we know ; then .-. A = 2mr tan - Thus we have two = -4 tan-=±2' of the angle of equi-cosinal angles. group -cos^ _ _1 ~iTcosli~rB~4' 1 A is that it must be one of a group Let a be the smallest positive angle of this =t a. tan (rnr ± ^ j = tan ^ ±^ j = ±tan | values differing only in sign. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 301/467 8/12/2019 Elementary Trigonometry by Hall and Knight RLEMKNTARY TRTOONOArETRY. 25S [CHAV. When any one of tlio functions of an acute angle A is we may in some cases conveniently obtain the functions 262. given, of A — , as in the following example. Given cos Example. ^ Make a in b A =a QR = b; cos A — to find the functions of right-angled triangle which the hypotenuse base . . i VQH PQ — a, and then PQR ^^ = ~ = cosA; =^ PQ a iPQR = A. RQ to S making QS = QP; :. Produce PSQ= L .: :. R A 1 L S-P<? =.2 ^ PQ^=2 PS^ = (a + hf + {a^-h^) = 2w' + 2ah\ :. The functions From written down in terms of the PRS. Art. 125, cos it PS=^2a(a + b). - may now be of sides of the triangle Thus A SR = a + b, and PR = Ja^-h^, Now 263. Q S A=A cos^ that appears equation to find cos we have A ^; if A s^ o cos A - 3 cos ^ ' . o J be so that cos A - given we have a cubic has three values. Similarly, from the equation sin it ^ = 3 sin ^ - 4 sin^ ^ appears that corresponding to one value of sin three values of sin A there are . be a useful exercise to ])rove these two statements analytically as in Arts. 254 and 257. In the next article we It will shall gix'c a geometrical explanation for the case of the cosine. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 302/467 8/12/2019 Elementary Trigonometry by Hall and Knight FTTNOTTONS OF XX.] (liven fos 264. A to find i-os '— , and 2n9 ANnLER. STTniMITLTIVT.F. explain the presniee of to the three values. Let a be the smallest positive angle with the given cosine then A=^2nn±a, and we have ; to find all the values of cos- (2w7r±a). Consider the angles denoted h\ the formula (27i7r and ascribe the values to + «), n in succession 0, 1, 2, 3, When M = 0, the angles are ±^, bounded by OPj and OQ^ %= 1, the angles are ^ - 1 bounded = 2, the angles are -|^ ±^, bounded by OP^ and when when By ?; giving to n the values 3, ' 4, 5, ... we with those indicated in the by OP., and ; OQ.^ OQ^^. obtain a series of figure. angles coterminal Thus OP,, 0Q„ OP^, 0Q.„ OP^, OQ, bound all the angles included in the formula -{2,mr±a). Now cos XOQi = cos XOP, = cos COSZOP3 = COSZO$2 = COS ^ ; /2n( 3 - a\ ; gj /27r cosZ0§3=cos X0P.2 = cos \r^ + Thus the values of A cos - a are cos - , http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight ^ +a —— 2ir cos a ^ , cos in^ . 303/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELKMENTARY TRIGONOMETRY. 260 EXAMPLES. 1. If ^ = 320°, -l + Vl + tan^J A tan 2 Shew that ^ l A= . tan 3. b. prove that A_ 2. XX. Find tan J — + Vl+tan2 24 when 1 , . .1 ^r-. tan 2^1 when cos 12 2^=— when cos .1 A and = ^,^^„ 110. hes between 180° 225°. and 4 Find cot A = - 4 ^ 5 2 and A hes ])etween 180° and 270\ 5. If cot 2^ = cot 2a, shew that cot ^ has the two vahies cot a and -tana. 6. Given that sin ^ sin 7. If tan ^ J, a 8. sin , = tan a, tan „ = sin a, shew . , tan Given that cos 3^ shew that the vakies of IT— a --— . TT +O 3 the values of tan - are tliat +o -— — - sin . sin r are IT — - tan -—- TT , , = cos 3a, . shew that the vahies of sin ^ are + sina, 9. Given that sin -sin(- + al, 3(9 = sin 3a, sinl — ±a). shew that the vahies of cos^ are + cos a, cos f ^ ±a j , http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight cos ( ^ ±a V 304/467 8/12/2019 Elementary Trigonometry by Hall and Knight CHAPTER LIMITS XXI. AND APPROXIMATIONS. If 6 he the radian measure of an angle less than a right angle, to shew that sin 6, 6, tan 6 are in ascending order of magnitude. 265. Let the angle 6 be bv repre.sented A OP. With centre and radius OA describe a circle. Draw PT at right angles to OP join to meet OA produced in T, and PA. Let r be the radius of the £ÂOP=\a0 .OP?.in Area of AOP=- area of .sector area of r-6 . OTP AOP=\r'^f^me^, ; AOPT=l OP PT= ^ But the areas triangle circle. of the triangle r . AOP, -r% are in ascending order of magnitude H. sin =\ r- tan the sector are in ascending order of magnitude -r^sin^, .•. r tan 6 d, 6, 3. AOP, and ; that the is, -rUan^ ; tan S are in ascending order of magnitude. K. E. T. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 19 305/467 , 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 262 When 6 266. indefinitely diminished, is — — each have unity for and -z [OHAP. prove that to sin 6 6 their limit. In the last article, we have proved that sin 6, 6, tan 6 are in Divide each of these quantities ascending order of magnitude. by sin ^ ; then 1, ' that 6 — —. sin 1 - — —. is, ' is 1 ; ^ between lies and sec 1 sin 6 But when 6 is are in ascendina; * * order of matinitude cos ^ (9' hence the limit of —. —6 v is sm fi. the diininislied, indefinitely is 1 that ; ; is, limit of sec ^ the limit of 6 — unity. Again, tan 0, we nitude. l)y dividing each of the quantities sin find that cos 0, Hence the a , tan limit of 1 —-— . sm e - J - is indefinitely - :. Similarly - sin - .6 . sin - n 6 ?i but since n . = when ?) =x . \ n the limit of sin Lt. ( ( msin- n tan http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight ) | . e 0\ sin - -^ - ^f p small, Lt. the forms J e=o\ n „ —6 in Zt.(^-^\î. Find the limit of n n by imity. is Lt.(^^^\^l, Example. tan are in ascending order of mag- These results are often written concisely e=oV B, 6, ; nj — - is unity; n n =- = d. = 6. 306/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight AND APrROXIMATIONS. LIMITS XXT.] '203 It is important to remember that the conclusions of 267. the foregoing articles only hold when the angle is expressed in radian measure. any other system of measurement If is used, the results will require modification. Find the value of Example. Lt. ( ^ ) « M=0 \ Let d be the number of radians in e —n^7:=-, lot) . and 71 = TT is w sin ^ • sin 6 180' 1806 n n . IT sin n° When n Rmjro — sm^; , also ; TT then if\ 1806' / t) ' indefinitely small, 6 is indefinitely small; n „=oV J 180* 9=0 V & /' /sin n°^ ..Lt.{ ;'=oV~^r j-jg^ 7-i8o 268. When we have shewn S is the radian measure of a very small angle, that sin^ that is, ^=1, cos^=l, sin^=^, cos^ = . , ^ l, tan^ ^^ , =1; tan^ = ^. tan d = rd, and therefore in the figure of Art. 265, the is very small. tangent PTîs equal to the arc FA, when z Hence /• AOP be shewn that these results hold so long When this is as 6 is so small that its square may be neglected. the case, we have In Art. 270, it will sin (a + 6) = sin a cos d + cos a sin = sin a + d cos a cos (a ; + ^) = cos a cos ^ - sin a sin ^ = cosa — ^sin H. 19—2 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 307/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight \CHAV. KTiEMENTAUY TRTfiONf iMKTRY. 2fi4 Example OA Let and PN inclination of a railway to the liorizontal plane feet it rises in a mile. The 1. is 52' 30 , find how many be the horizontal plane, OP a mile of the railway'. perpendicular to OA PN = X Let lPON=0; feet, PN then ^ = radian =6 sin 6 OP' But Draw measure of approximately. 52' 30 52^ ^ = 180 bO 22 1760 X 3 Thus the rise is 80» 8 7 TT J_ ^ IT 8 ^ l80 ^ _ 186' 7 1760 X 3 X 22 242 8x180 3 = 80f. feet. Example 2. A pole 6 ft. long stands on the top of a tower 54 ft. high: find the angle subtended by the pole at a point on the ground which at a distance of 180 yds. is A be Let BC ground, the the from the foot of the tower. point on the tower, CD the pole. Let then lBAC = a, iCAD = 6; 54 1 BC tan a AB 540 10' , , tan (a But tan(a + 0) = + tan a 1 .. ^) — —= = BD = + tan - AB 60 tan a 6 1-6 - tan a tan 6 1 10 1-16 whence 6 contains On = — - 91 that , x is, the angle + tan a approximately; m + 10-6 i ^ 1 - ' 1 is ^ of a radian, and therefore degrees. TT reduction, we find that the angle is 37' 46 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight nearly. 308/467 8/12/2019 Elementary Trigonometry by Hall and Knight LmiTS ANiJ XXI.] If 6 be the number of radians in 269. cos^>l — - that an acute and «iii^>^ , 4 /iio-9^ — z sin- - cos^= oiiicc ani/(e, (o jj?'ove . 2 i_>- 265 Al'l'HOXlMAl'IONS. 1 ,.66 and sin-<-; , C08^>l-2.^ .-. 01 that cos is, Again, sin ^ 6>\—~. 6 = 2 sin - cos tan g but .-. •. But sin g <g , sill 6 „ = >^ ^>2 6 6 2 tan - cos^ - ; : -COS-- 2 sin<9>(9 2 1 -sin--j, and therefore .-. iim6>6- — . 4 270. From follows that if 6 the jn-opositions estabhshed in this chapter, is an acute angle, cos 6 lies between and it sin 6 lies Thus cos 6 = 1 ~ proper fractions less 1 and 1 — between 6 and ^ — 6^ -h j 6^ -r • 4 and sin ^ = ^ — X-'^^, where k and than ^ and | respectively. is- http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight /•' are 309/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 310/467 8/12/2019 Elementary Trigonometry by Hall and Knight AND LIMITS XXI.] -._-. „, - U Sill , • —^- lo shew that 272. 7t 7 continually decreases 6 continually increases from We 267 Al'l'UOXlMATlUNS. — to J' jrom I i to - as . shew that the fraction shall first sm(d + h) sin 6 ~6 hh- . ^«P««itive, k denoting the radian measure of a small positive angle. „„ ,. ,. . i his Iraetion ^ . _6s,\nd{\.- cos h) ^ Now tan ~ 6>B^ that . Also 1 - cos k is — A + cos^sin/t) + /i)sin^-^(sin5cos — (^ = • . , . ,t {0-\-h) , + (A sin ^ - 6{e^h) > ^ cos 6, and sin 6>6 cos 6 sin sin ^ is h ^ cos 6 sin h) h > sin h ; /*. positive and therefore the fraction is ; hence the numerator positive is positive, ; + h) sin 6 e+h ^ ~T sin (6 •'• .•. —-— continually decreases as d continually increases. wi, Wnen a d sin — ~— = 1; = A0, (9 Thus the proposition ^ is , r In . this TT , and when 6 = -, sin(9 — = tt 2 • established. EXAMPLES. XXI. » • Exercise take n- a. = 22 ^ A tower 44 feet high subtends an angle of 35' at a point find the distance of A from the tower. on the ground 1. A ' : 2. From the to[) of a wall 7 ft. depression of an object on the ground from the high the angle of 24' 30 : find its distance 4 is in. wall. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 311/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETKV. 268 [CHAP. Find the height of an object whose angle of elevation at a distance of 840 yards is 1° 30'. 3. - «^ Find the angle subtended by a pole 10 4. ft. 1 in. high at a distance of a mile. 4\ Find the angle subtended by a circular target 4 diameter at a distance of 1000 yards. 5. feet in Taking the diameter of a penny as r25 inches, find at what distance it must be held from the eye so as just to hide the moon, supposing the diameter of the moon to be half a 6. degree. 7. Find the distance at which a globe subtends an angle of 5'. 8. Two 9. A man 1 1 inches in diameter the same meridian are 1 1 miles apart find the difference in their latitudes, taking the radius of the earth as 3960 miles. 120 ft. : places on : high stands on a tower whose height is shew that at a point 24 ft. from the tower the man 6 ft. subtends an angle of 31 'o' nearly. A flagstaff standing on the top of a cliff 490 feet high 10. subtends an angle of -04 radians at a point 980 feet from the base of the cliff find the height of the flagstaff. : 11. When 71 = 0, find the limit of (1) 12. When \Vln;ii ^ ^„ li — «i^'. ?i^; (2) 1 ao, find the limit of - nr 2 = 0, 7n aiu ' ' = -01 15. If ^ 16. Find 17. (Jiven tlie — 27r n . find the limit of 1— cos^ 6 Bind sin md — nam n() ta.nm6 + ta,nnd of a radian, calculate cos ' l^ + d) vnhie of sin 30° lO; 30 . CO.S ( ; -|-^ j = -49, find the sexagesimal \;due of d. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 312/467 8/12/2019 Elementary Trigonometry by Hall and Knight DISTANCK AND DIP XXI.] mu THK IIOIUZON. Ol' Distance and Dip of the Visible Horizon. A Let 273. earth's surface, be BCD a AE cut the From A draw BCD centre E circumference in B lFAC AC to AF at right touch the angles to called the dip of is its circle and join BC. in C, Draw the section of the earth by a plane passing through and A. Let and D. above point a as seen from A. Thus the dip of AD then ; the horizon the horizon is the angle of depression of any point on the horizon visible from A. To find the distance of the honzon. 274. In the figure of the last then by Euc. that AB = h, iii. 36, is, AC=x; EBÊD = AC'^ÂB AD r, . a.-2 ; ; = h (2?- +h) = 2hr + h:\ For ordinary altitudes 2/ir article, let h ^ is very small in comparison with hence approximately x'^ In miles, let \j'2hr. suppose the measiu'ements are made then a be the number of feet in AB taking r = in ; a By and x = formula, this and = 2hr = 1760x3xA. we have ;396'0, „ 2 X 3960 X a 3a 1760x3 we have the following rule : Thus Twice the square of the distance of the horizon measured in miles is equal to three times the height of the place of observation measured in feet. Hence a man whose eye is 6 feet from the ground can see to a distance of 3 miles on a horizontal plane. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 313/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 27U [chap. Example. The top of a ship's mast is 66f ft. above the sea-level, After the and from it the lamp of a lighthouse can just be seen. ship has sailed directly towards the lighthouse for half-an-hour the lamp can be seen from the deck, which is 24 ft. above the sea. Find rate at the Let L the ship denote the lamp, positions of the ship, G is is sailing which B D and E the two the top of the mast, the point on the deck from which the lamp is a tangent to the earth's seen; then LCB surface at A. [In problems like this must some of the lines necessarily be greatly out of proportion.] Let AB and AC he expressed in miles; then since DJ5^66| feet and EC = 24. feet, we have by the rule JB2=|x66t::=100; .-. ^B= 10 ylC2 = .•. |x 24 AC = & But the angles subtended by miles, = 36; miles. AB ^C and at the centre of the earth are very small .-. Thus the ship 275. horizon ; AD = AB, and arc AC = AE. DE = AD - AE-AB - AC = A mi\es,. [Art. 268.] arc .-. &XC sails 4 miles in half-an-hour, or 8 miles per hour. Let 6 be the number of radians in the dip uf then with the figure of Art. 273, we ha\-e . cos d EC = -f^-r = .-. 2810 2 Since 6 and - are small, r f-— tlie i + = f,h\-' = r - i-f... r- we may neglect the terms on the right after the http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight replace sin 7; by - and first. 314/467 8/12/2019 Elementary Trigonometry by Hall and Knight DISTANCE AND XXI.] ^= il'i''^ Let ^V' Now DIl' j.^ OF THK HOIUZON. «• ^=\/t- be the muubcr of degrees ^'r=6'S nearly ; 271 180^ 180 n TT in 6 radians ; then /U \ ' r hence we have approximately 180 X 7 X \/27i 22 X 63 ' #=^V2A, or a formula connecting the dip of the horizon in degrees and the height of the place of observation in miles. EXAMPLES. Here 22 ;r = -zr and , XXI. radius of earth b. = .3960 miles. Find the greatest distance at which the lamp of a lighthouse can be seen, the light being 96 feet above the sea1. ^ level. 2. If the lamp of a lighthouse begins to be seen at a distance of 15 miles, find its height above the sea-level. 3. The tops of the masts of two ships are 32 V^ 8 ft. in. and / ft. 8 in. above the sea-level find the greatest distance at V which one mast can be seen from the other. 42 : Find the height of a ship's mast which is just visible at a distance of 20 miles from a point on the mast of another ship which is 54 ft. above the sea-level. 4. the mast of a ship 73 ft. 6 in. high the lamp of a lighthouse is just visible at a distance of 28 miles find the height of the lamp. 5. From : 6. a, hill Find in minutes and seconds the dip of the horizon from 264U feet high. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 315/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 316/467 8/12/2019 Elementary Trigonometry by Hall and Knight CHAPTER XXII. GEOMETRICAL PROOFS. 276. To find the expansion o/tan {A Let z.ZOJ/=J, and In to to OL OL lMON=B; +B) geometrically. LLOX=A->rB. then ON take any point P, and draw PQ and and Oi/ respectively. Also draw RS and and PQ respectively. RS OS 1- * PT OS TR RS also the triangles PT OS'^ OS ^_TR TP' TP 'OS Now PR perpendicular RT peqiendicnlar • O'S TR -y^, =tan A, and ~^=tan ^4 ; ROS and TPR are similar, and TP PR = tan B. OR OS tan A + tan B ta.n{A + B) = 1 - tan A tan B http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight therefore 317/467 8/12/2019 Elementary Trigonometry by Hall and Knight KLEMENTARY TRIGONOMETRY, 274 manuer, with the help of the figure on page 95, we obtain the expansion of tan (A — B) geometrically. Ill may [chap. like 277. To prove geometrical^ the formulce for transformation of sums into products. Let L EOF be With centre meeting 00 in H denoted by J, and L by B. and any radins describe an and OF in K. L KOH by OL Bisect EGG ; then OL bisects arc of a circle HK at right angles. LR perpendicular to OE, and through L draw OE meeting KP in M and QH in N. Draw KP, HQ, MLN parallel to MKL and are KM=NH, ML = LN., PR^RQ. NHL It is easy to prove that the triangles equal in Al«o all L. respects, so that GOF= A - B, and therefore lHOL= lK0L = lEOL = B + A^ n J<^P HQ A-B A-B A+B 2 KP + HQ {KM+ LR) + {LR - NH) _ ^ LR ~ OK' OK http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 318/467 8/12/2019 Elementary Trigonometry by Hall and Knight TRANSFORMATION OP SUMS INTO PROPTTOTS. XXTT.] ... 275 sin^ + sm5 = 2^.^=2sin/;<9ZcosA'0Z ^ = A-B A+B . —— sm 2 - cos —^ . OP+OQ OF OQ _ cosJ+eo,s^=^^ + ^^=-^^— _ OR-PR) + {OR+RQ ^g OJl ÔK OK ) { , . sm .-1 =2 OL — =2 COS „ -sm « = . 2 cos 7?0Z CO.S rOZ ^,= OK . A+B COS A-B —— ^— . KP-HQ 5^^-— ^ A'P ir$ - (Xi2 - iVAT) _ - {KM+LR) OK ^ ^J/ OK = 2^ . J^,=2 cos ZA'J/sin KOL KL OK = since ^ ^ 2 cos A+B — — sm A-B —^ /.A'J/=comp' of == lKLM= OP OQ , . - , 1-1-7? l ML0= lL0?:='—^. OQ-OP (OR+R Q)-{OR-PR) „PR_^ML 'ÔK - OK OK = 2 ^^ ^=2 KL OK ^ . = 2sm . A+B 2 . «' sin L KM Bin KOL A-B -2~- http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 319/467 8/12/2019 Elementary Trigonometry by Hall and Knight KLKAFKNTARY TRIflOXOAfKTRY. 276 Geometrical proof of the 278. BPD Let the diameter, On farimihi BD be a semicircle, C Â [fHAP, the centre. the circumference, take any point ]\ and join Draw PB, PC, PD. PN perpendicular to BD. lPBD = A, Let then N D lPCI>=2A. And . sin I. ^ v. XPB = com])' of L PDN= — L PBD = A. PiV 2PN 2Piy „ PJS^ BP =— = = =: 2 2CP BD BP' BD CP , A - = 2%m PBN cos PBD = 2 sin A cos %A = cos ^4. GN WN CN+GN CP BD BD BN- ND BD {BN- BC) + {CD - ND) BD BN BP 'BP BD ND PD PD- BD ' = cos A . cos A - sin A . A sin = cos2^ — sin^ J.. '* * ^^ ,CN_ CD-DN CP t'P ^ DN _ 2DN BD CP nN DP ^ ^ . . . . :l-2sin2J. cos 2 J. : CN CP ~ BN-BC CP =2 BN CP' BN BP_ = 2 cos A BP BD 1 ' 2 cos2 2BN BD -1 1 . cos A A -1. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 320/467 8/12/2019 Elementary Trigonometry by Hall and Knight VALUE OF SIN XXIX.] tan 2A 18° 2FN 2Fiy BN-ND PN By : BN ND 1BN 1 ^ — tan A yj) PiY PN BN A 2 tan 1 277 . tan A tan^ l-tan2^ 2 279. To find o/sin 18° geometrically. the value Let ABD be an isosceles triangle in which each angle at the base BD is double the verthen tical angle A ; .1+2.1+2.1 and therefore Bisect BD .1 = 36°. lBAD .-. where From AE by at right angles Thus = 180°, then ; AE bisects 3 BAE=18\ L sin 18° AB = a, = ^= AB and -, a BE=x. the construction given in Euc. iv. 10, Aa=BD = 2BE=2x, AB.BC=AC-'; and ' The upper .sign .•. a(a — 2x) = {2xy; .-. 4r2+2a.r-a2 = 0; ^'= -2a + \/20a2 -l-k-Jb 4— =r— =— — nuist be taken, since sin 18' .f is • positive. Thus = ^^. U. K. E. T. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 20 321/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 378 Proofs by Projection. Definition. 280. from any If A and B, lines AC and drawn perpendicular to OX, two points BD are then the intercept CD called is the projection of .1^ upon OX. E parallel to OX then CD = AE=ABgo9BAE; CD = AB cos a, Through A draw A that is, where a is ; the angle of inclination of the lines AB anel OX. To shew that the projection of a straight line is equal to projection of an equal and parallel straight line drawn from a 281. the fixed point. Let <) AB be any straight line, a fixed point, which we B shall call the origin, line OP atostraight equal and parallel AB. Let CD and OM be the pro- AB and OP upon any line OX drawn througli jections of straight the origin. Draw A E })arallel triangles The two OMP are identically to 6'A'. AEB equal .-. that is, projection and Q OM=AE=CI)of (>/* = projection http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight of AB 322/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight PROOFS BY PROJECTION. XXII.] 279 Hence OP denotes a line drawn from the origin parallel to AB, and OQ denotes a line drawn from the origin parallel to BA Similarly the direction of a projected line fixed is l)y the order of the letters. Thus Avhile CD DC is Hence drawn to the right from C to drawn to the left from D to C and is in sign as well as in OM=CD, that is, and 1) and is negative. is positive, magnitude and OX=DC; projection of OP = projection of AB, projection of 0$= projection of BA. Thus the projection of a straight line can he represented both in sign and magnitude hy the projection of an equal and parallel straight line drawn from the origin. Whatever be the direction 83. of AB, the line OP will fall within P Y one of the four quadrants. Also from the definitions given in Art. 75, we have ~-<ôsXOP, that is, OM=OPcoiiXOP, whatever be the magnitude of the angle XOP. We shall always suppose, unless the contrary is stated, that the angles ai'e measured in the positive direction. Let 284. any two and Q points. QX perpendicular We P OP, OQ, PQ, and draw P3/ Join and be the origin, to OX. have OM=OX+XM, since the line negative ; XM that is N X to be regarded as is, the projection of 0/'=i»rojection of O(j + iivoiect\o\\ nf (Jl'. •20— http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 323/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONUMETRY. 280 Hence, the sum Tims [chap. projection of one side of a trianyle is equal to order. of the projections of the other two sides taken tlie m projection of (7^ = projection of Oi'+ projection projection of (i'P^ projection of <'^(> oi l'<^; + projection of 01', General Proof of the Addition Formulae. OX lu Fig. 1, let a line starting from revolve luitil it lias traced the angle A, taking up the position OM, and then let it further revolve until it has traced the angle B, taking up the 285. OX. final position Fig. XOX is the angle A+B. Fig.2. I In also Thus OX take draw OR Projecting any point P, and draw J'(^> perpendicular equal and parallel to (JP. upon OX, we have projection of .-. to OJI 0P= projection of 0^ + projection of (^P = projection of 0§ + projection of OR. OP COS XOP = 0Q cos XOQ + OR cos XOR (1) = OP cos B cos XOQ+ OP sin B cos XOR; . that is, . cos c;<js XOP = cos B cos XOQ + sin B cos XOR (A +B) = cos B cos A + sin B cos — cos A Projecliiuj (ipoii )', cos B - fjn Î wc have ojdy http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight (90° ; + A) sin B. to write J' foi' X in (1); 324/467 8/12/2019 Elementary Trigonometry by Hall and Knight 281 PROOFS BY PRO.IKrTTON. XXTI.] OP COS YOP = OQ cos YOQ + OR cos YOR = OP cos B cos YOQ + OP sin B cos thus J V^/i* • . that cos . YOP = cos 5 cos rO§ + sin 5 cos YOR ; is, {A+B cos .-. 90°) sin(^ = cos B cos (^ - 90°) + sin B cos J + 5) = sin^cosi5 + cos^4sin ; B. In Fig. -2, let a line starting from OX revolve until it has traced the angle A, taking up the position 03f, and then let it revolve back again until it has traced the angle B, taking up the is the angle A-B. Thus final position ON. XON In ON take any point P, and draw PQ perpendicular to also draw OR equal and parallel to QP. produced MO ; Projecting upon OX, we have as in the previous case P cos XOP = OQ cos XOQ + OR cos XOR = OP cos (180° - B) cos X0$ + OP sin (180° . that • . cos .TOP = - cos 5 cos .TO§ + sin - 5) cos 5 cos XOR XOR ; : is, cos (J -B)=-cosB cos (4 - 180°) + sin 5 cos = -cosB{- cos J) + sin B sin A = cos ^4 Projecting upon Y, cos B + sin J (.-1 - 90°) sin B. we have OP cos TOP = OQ cos F0§ + OR cos IÔii = OP cos (180° - 5) cos YOQ + OP sin (180° ; . • . that cos cos - B) cos TOT? ; FOP = - cos B cos YOQ + sin 5 cos YOR is, (A-B- 90°) = .-. - cos i? cos (.-1 - 270°) + sin 5 cos (J - 1 80°) H\n(A-B)= -cos5(-sin^)+sin5(-cos.4) = sin A cos B - cos A http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight sin B. 325/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 282 The above method 286. of proof is [CHAP. appHcable to every case, and therefore the Addition Formula? are universally established. The universal truth of the Addition Formulae may also be investigations of Arts. 110 deduced from the special geometrical and 111 by analysis, as in the next article. When 287. each of the angles A, B, A +B m than less 90°, we have shewn that cos But cos {A ( J. + i?) = cos ^l cos B-s,\\\A sin + B) = ^\n {A+B + also 90°) = sin ^ = sin {A + 90°), - sin A = cos (A + 90°). (.-1 5 +90° (1). + B) ; cos and Hence by substitution sin {A + 90° + B) In like manner, cos (4 = sin in (1), (A we have + 90°) cos B + cos {A + 90°) sin B. may be proved it [Art. 98.] that + 90° + J5) = cos (4 + 90°) cos Z?- sin (4 + 90°) sin B. Thus the formulae for the sine and cosine o{ A+B hold when A is increased by 90°. Similarly we may shew that they hold when B is increased by 90°. repeated applications of the same process it may be proved that the formulae are true when either or both of the angles A and B is increased by any multiple of 90°. By = cos A cos ^-sin 4 sini5 (1). +B) cos (4 + B)=- sin {A+B - 90°) = - sin (4 - 90° + B) cos {A Again, But ; 4 = - sin (4 - 90°), sin 4 = cos (4 - 90°). cos also and Hence by substitution sin in (1), (4-90° + B) = sin (4 - [Arts. 99 and 1 02.] we have 90°) cos we may shew that (4 - 90° + j5) = cos (4 - 90°) B + cos (4 - 90°) sin B. B - sin (4 Similcirly cos Thus the formulae A is aiT for the sine diruinished by 90°. trill' wlieii /i is In like cos and cosine of - 90°) sin B. 4 + /? hold when manner we may prove that they diminished by 90. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 326/467 8/12/2019 Elementary Trigonometry by Hall and Knight MISCELLANEOUS EXAMPLES. XXII.] 283 H. repeated applications of the same })roces.s it may be shewn that the formiilee hold when either or both of the angles A and B is diminished by any multiple of 90'. Further, it will be seen By that the formulae true are either of if increased by a multiple of 90 multiple of 90 A the angles and the other or B is diminished by a is '. = >iin Poos Q + cos Psin cos {P+Q) = cos P cos () - sin P sin Thus .sin and P=J±m.90°, where VI [F+(^) and n being any positive and Q, Q, Q=B±n.Q(r, integers, and A and B any acute angles. of Thus the Addition Formuke anv two angles. are true for the algebraical siun MISCELLANEOUS EXAMPLES. If the sides of a light-angled triangle are 1. cos 2a H. + cos 2/3 + 2 cos (a + /3) shew that the hypotenuse If the in-ceutre 2. is and 4 cossin 2a —— + sin 2/3+2 sin (a + /3), . and circum-centre be at equal distances from BC, prove that cos5 + cosC=l. The shadow of a tower is observed to be half the known height of the tower, and some time afterwards to be equal to the height how much will the sun have gone down in the interval ? Given log 2, 3. : Ztan 4. If 63° 26' = 10-3009994, diff. for l' = 3159. (]+sina)(l+sin/3)(l+siny) = (1 shew that each expression Two is equal to -sina)(l -sin/3)(l -sin-y), +cosa cos/3 cosy. on the same side of the centre subtend 72° and 144° at the centre prove that the distance between them is one-half of the radius. 5. parallel chords of a circle lying : Also shew that the sum of the squares of the chords to five times tlie square of tl>e radius. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight is e(|ua] 327/467 . , 8/12/2019 . Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 284 Two [CHAP. an angle of GO . From their point of intersection two trains P and Q start at the same time, one along each line. P travels at the rate of 48 miles per hoiu , at what rate must Q travel so that after one hour they 6. straight railways are shall be 43 miles apart shew that = co.s-i- + co.s-if sin^ a =— ,- ab a'' If/', 8. at ? a If 7. inclined cos a + rr, b' r denote the sides of the ox-central ti'iangle, i>rove (J, that je a^ A g2 2abc , formed by two straight roads OA and OB, and subtends angles a and /3 at the If OA = a, points A and B where the roads are nearest to it. and OB = b, shew that the height of the tower is 9. tower is Va^ - within situated b^ sin a sin /3 / the ^/smia ->,- angle sin (a /3) - ^). In a triangle, shew that 10. ,.2 If 11. AD he + n^ + a. ,.,,2 + ,.^2 = 1 6/?2 _ a'- - Z>2 - c2. median of the triangle ABC, shew that (1) cot 5.4 Z) (2) 2 cot =2 cot ^+ cot 5; J DC= cot B - cot C. r are the distances of the orthocentre from the 12. sides, If p, q, prove that a h €\ fa b c\ lb c P q >-)~\P 9. r) \q r a\ tc a b p) p q \r Graphical Representation of the Circular Functions. Definition. Let f{x) bo a function of x which has a single value for all values of .r, and let the values of x be represented by lines measured from along OA' or OX', and tlio values of /(.f) by lines drawn perpendicular to XX'. Then witli represent any value of x, the figure of the next articlii, if and the corresponding value of /(.r), the curve traced out by the point P is called the Graph of /(.)•). 288. MP OM http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 328/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight AND rOS OliAPHS OP SIN 6 XXII.] Graphs of sin 9 and cos 285 0. 9. Suppose that the unit of length is chosen to represent then any angle of 6 radians will be represented by a a radian 289. ; line OM which units of length. contains Graph of sin 6, Let MP, drawn perpendicular to OX, represent the value of corresponding to the value of then the curve traced sin out by the point P represents the graph of sin 0. OM As to 1, OM or which increases from is its OM increases from As OM to As from — As OM 1 to OM Stt, , OM - , 3/P or sin increases from greatest value. As from to - ; increases - to n, from tt to MP decreases from -— MP increa.ses MP decreases nmnerically , to 1 numerically 1. increases from — to -Itt to 27r, 0. increases from 47r, from 3fP passes through the same increases from Since sin to 4ir to Gir, from series of values as G/r to when 2jr. — 0)= - sin MP the values of lying to the left of are equal in magnitude but are of opposite sign to values of MP ( 0, lying at an equal distance to the right of 0. Thus the graph a continitons waving line extending to an infinite distance on each side of 0. The graph licing at of sin ^ of cos^ is is the same as that of sin the point marked - in the ^, the origin figvu-e. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 329/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 286 Graphs of tan 2S0. As [chap. and cot 9. 6 before, suppose that the unit of length is chosen to then any angle of 6 radians will be represented by a line OJ/ which contains 6 units of length. I'epresent a radian ; Let AfP, drawn perpendicular to OX, represent the value of tan coi-responding to the value 03f of then the curve traced out by the point F represents the graph of tan 6. ; By tracing the changes in the value of tan ^ as ^ varies from from Stt to 47r, it will be seen that the graph of tan 6 consists of an infinite number of discondmious equal branches as represented in the figure below. The part of each branch beto 27r, , neatli XX' is convex towards XX', and the part of each branch above XX' is also convex towards XX' hence at the point where any branch cuts XX' there is what is called a point of inflexion, wliere the direction of curvature changes. The proof of these statements is however beyond the range of the present work. ; The various branches touch the dotted the points marked 'in at an infinite . ~ 2 ^r distance from XX'. Graph lines passing through 5 2 of tan 0. X' The student* should draw the graph similar to that of tan of cot (9, which is very 0. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 330/467 8/12/2019 Elementary Trigonometry by Hall and Knight Graphs of sec 291. AND COSEC GRAPHS OF SEC XXII.] and cosec The graph It consists of an of need in finite 281 6. 0. is re^jresented in the figure below. number of equal festoons l}'ing alter- nately above and below A'.Y', the vertex of each being at the The various festoons touch the unit of distance from XX'. dotted lines passing through the points marked 2' z' ~2^ ' at an infinite distance from XX'. Grapli of sec The graph of cosec ^ is being at the point marked 0. the same as that of sec - 6, the origin - in the figure. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 331/467 8/12/2019 Elementary Trigonometry by Hall and Knight CHAPTER XXIII. SUMMATION OF FINITE SERIES. An expression in whicli the successive terms are formed by some regular law is called a series. If the series ends at 292. some assigned term terms is A unlimited may series called it is and term, M„ = cot2 if number of an infinite series. obtained from m„, the is Un = G0s{a + 7i^), then Un + if the if ; be denoted by an expression of the form where tin + i, the (n + l) by replacing nhy n+l. Thus called a finite series it is ia, then ?/-„ i + = cos{a + (n+l) fi} j = cot 2 term, n ; a. term of a series can be expressed as the difference of two quantities one of which is the same fvxnctioxi of If 293. the r r that the other is of r + 1, the sum of the series may be readily found. For let the series be denoted by Ui and its sum by >9, + U2 + U.^+ and suppose that any term U/.jt thou ,V= (v.^ Example. - Vj) + (v.^ - v^) -^ + cosec cosec 2a + {v^ - a=. sin a + ~~ t/*. J. 1 Find the sum of the cosec a +M„, v.^) Vf J + .., + (v„~v,,^{) + (v„ + - v„) 1 series cosec 4a + + cosec ( 2 = ^—^^.— . sni a . sm . a http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight sni 2 ~' a. i) a ^ . sm a 332/467 8/12/2019 Elementary Trigonometry by Hall and Knight SUMMATION OF Hence If we cosec a cosec 4a = cot '2a - cot -la, '2 -i the a - cot '2a. a^cot 2 -â - S' To find a. = cot addition, 294. - cot cosec 2a cosec By a — cot v we obtain leplacu a by 2a, Similarly, 289 I'UNITK SERIES. — cot ^ sum of the cot - cot 2 a. â. of a si)ies 'i''^' of n aiujles which series are in arithmetical progression. Let the sine-series be denoted by sina + sin(a+/3) + sin(a + 2/3)4- We . 2 sin (a 4-/3) sin 2 sin (a By {a + (m- )/:{}. have the identities 2smasin 2sin + sin lo -t- 2/3) + (M-l)i3} sin 0\ = cos / a-^j ;3 - I ( = cos (n+^) | = cos (a + 8 sin'- / = cos( ~*^*^^ &\ a-H^ I *^^s ( ) ( , ^T/' y j - cos (« + y j 2n-S — aH ^ \ /3 - cos 1 ( ( > 2m -1 a-f---^ . P addition, & 2A sui ^ = cos / f a - /3\ ^ 1 cof; / I a+ —-— 2;i-l n-\ \ = 2sin(a-|-— /3jsin- «./3 . ^ 2 ^S=—— . sni / ( n ^- + ^— )3 1 ^. sin - http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 333/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 334/467 . . . 8/12/2019 Elementary Trigonometry by Hall and Knight 291 SUMMATION OF FINITE SERIES. XXin.] Some 298. series be brought under the rule of Art. 296 may by a simple transformation. Example Find the sum of n terms 1. of the series cosa-cos(a + /3) + cos(a + 2|3)-coo(a + This series COsa + 3/3) equal to is + cos(a + i3 + 7r)+cos(a + 2j3 + 27r) +COS + (a 3/J a series in which the common difference of the angles last angle is a + («- ) 03 + t). + 37r) is /3 + + , tt, and the sm—^^^ S .•. ^^ cos = + , -.a a • -I (;i-l)(/3 + 7r) 2~ Example sin a Find the sum of n terms of the 2. + cos (a + /i) - sin (a series + 3/3) + sin + 2/i) - cos sin(a + 2/i + 7r) + sin( (a (a + 4^) + . . . Tliis series is equal to sina + sinf a + /J + ^ j + a + 3/3+YJ + ' IT common a series in which the H(2/3 . 111 sin ^ 2)3 sm 1. eacli of + ;.^ sin |_ -Ia + (;t- l)(2/j + 7r) , | ' , i XXIII. a. the following series to n terms sin u 4- sin 3a ^ + - -1 EXAMPLES. Sum is + 7r) —f_— 4_ ^ ,S'^ difference of the angles : + sin 5a + COs(a-/3)+COs(a-2/ii)+ 2. cosa + 3. .sin 4. cos a TT -J + sin a-- j+sin(a ( + cos 27r , + ÔS , -r- 3n-r- -r . http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 335/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TllIGONOMETRY. 292 sum Find the of each of the following series 5. COS-+COS— +COS— + 6. C0S- + C08- + C0S-+ 7. siu 8. COS - 9. sin na + + sin + cos n Sum n h sill n h COS n . to h /i - , 1 ,2-. , terms. to 2;i - 1 terms. 1- n + sin {n-l)a + sin (n -2) a + to 2;i terms. ....... each of the following series to n terms - + sin 3^ -sin : + 10. sin 11. cosa-cos(a-^) + cos(«-2/5)-cos(a-3^)+ 12. cos a 13. sin 20 sin 6 + sin 3<9 14. sin a cos 3a + sin 3a cos 15. sec a sec 2a + sec 2o sec 3a 16. cosec 6 cosec 3^ (9 - sin : '^'i«i9- +co« .7r.27r.37r, n [c'HAP. 2<9 sin (a - ^) - cos (a 5.t + sin 4(9 3(9 + + sin 5a cos 7a + + sec 3a sin sec 4a sec a + tan t;u I ^ 18. cos 2a cosec 3a 19. sin 20. into -^ sec - + cos Hu + 3d cosec 50 + cosec 17. . + sin (a -3/3) + -2^) sin 26 + cosec 4(9 i')d cosec Id + + tan -^ sec —g + cosec 9a + cos «sec3a + sin 3asec9a + sin 1 9a sec 8a cosec 27a + 27a+ The circumference of a semicircle of radius a is divided Shew that the sum of the distances of the n equal arcs. several points of section from either extremity of the diameter is a cot \ 4n From the angular points diculars arc drawn to A'X' and 21. 1 W of a regular polygon, perpen- the horizontal and vertical shew that the algebraical diameter of the circumscribing circle sums of each of the two sets of perpendiculars are equal to : zero. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 336/467 8/12/2019 Elementary Trigonometry by Hall and Knight SUMMATION OF FINITE SERIES. SSIII.] By means 299. 2 sin^ a of the icleutities = 1 — cos 2a, 4 sin^ a = 3 sin a we can sum find the 293 — sin 3a, 2 cos^ a = 1 + cos 2a, 4 =3 a cos'' cos a + cos 3o, and cubes of the of the squares and sines cosines of a series of angles in arithmetical progression. Example Find the sum of n terms of the 1. sin-a —n - I cos 2a + (2a siu iiB „ + cos 3 (cos a + - COS fa + (3 cos (2;( \ ; - ) + + (cos Sua + -:—5— sm 3a \ J no. 'Ja) sin 1) a) 3 sin +cos* (2?t- + cos 3a 2 ( cos na sin 4 sin a 300. 'I'lic ; of the series + cos 3ft + cos 5a + 3 sin na sin a 3a) } + -ij3}\ ^ , , Find the + 4/3) + + ' ' sum cos* a + cos-' 3a + cos* 5a + (8 cos a . , {1 - cos (2a -1)28} 2 )i - '2l3)\ + 2^) + cos (n + 2j8) + sic^(a 5^ 2. = (2a 2a +I2a + — ^ cos ^ smj8 Example i=: + cos sin iiB = 4/S— + sia2(a+/3) - cos 2a} + {1 - cos (2a + 2,S:= [1 „ + series (3 cos 1) a. 5a + cos 15a) I- 3a + cos 9a + cos 15a + cos ('Sa + ) (2n - 1) 3a) \ } 2 ( ; ) 3«a cos %na 4 sin 3a following further examples illustrate the principle of Art. 293. Example l As + Find the sum 1. 1.2.a;2 in Art. 249, tan 1 + ——— — ^.-, r (;h- 1) .•. K. 3. l+n{n + l)x^ a;- we have , 1 H. + 2. X ' of the series ,r- <S'-tan^' tanî (r-H ^ {n+ l).c - l)x- tau~'/u-; tau-^j;. K. T. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 21 337/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 294 Example Find the sum of n terms of the 2. tan a + AYe have Replacing a by .- tan - + tan a = cot ;^ ^5 tan ., and dividing by 2-3ta»i23+ we obtain 2, 2~ • lâlâl , , , , cot 2»i=i t**^ - t-, 2^ tan 2- = 2— — — J Similarly, By J + series a - 2 cot 2a. 2**°2 2 „. 22 2- 2^ = 2î^ a —-^ 1 S = On— :;-—r addition, ' a 2 cot 2 ^ ; 2^1 - 2^2 'ô* cot <ji EXAMPLES. — cot - 2 cot XXIII. + cos2 3^ cos2^ 2. siii -â 6. cos- « + cos- 4. sill 6 + sin^ 2<9 5. siii^ a + .siu^ \a-\ b. cos-* a + cos' 7. tan ^ + 2 tan + .sin2U + - j + / cos a + COS 3a 2(9 3<9 ) 27r\ 1 + 2^ L+— + cos- + siir (a sin ^ Ja ^\ ./• a ., • b. : + cos'^5^-|- 1. ., g^, 2a. Svim each of the followiug series to n terms „ [cHAP. + 27r\ + + sin'* + coS'* tan )-|-. ., 2-(9 ( a / (a + -- + ) 4fl-\ ) + . , + cos a + COS 5a http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight cos a + cos 7a 338/467 8/12/2019 Elementary Trigonometry by Hall and Knight SUMMATION OF FINITE SERIES. XXIII.] sin 26 siu- 9. + - siu- 20 siii 4^ + - sin- i a ft 4 295 4^ «in 8(J + fi 2 cos^ sin2- + 22cos-sin-—2 + 2^cos-2sin2-^+ 10. 11- ^^ ''t:^+*^ '^273W^+^^'^'^3-:^^+ 12- ^^^ 'r^^+^^'^-'iT¥T2-2+'^^^~'r+^+ 13- ^^'^ ^-T^+^^ ~'2+ii:2^+^^ '2-:fJT3*+---- tan-1—|-_p^, + tan-i^-^ + tan-Y_-^+...- 14. 15. any point on the circumference of a circle of radius From chords are drawn to the angular points of the regular inscribed polygon of n sides shew that the sum of the squares of the chords is 2'nf^. r, : From P within a regular pol}'goji of 2n sides, perpendiculars PA^, PA^, P^s, ...PA^n are drawn to the sides 16. a point : shew that PA, + PA^ + ... + PA,„_, = PA., + PA^ + ... + PA,„ = 7ir, where r is the radius of the inscribed circle. a regular polygon and P a point on the circum.scribed circle lying on the arc AÂ^n + n shew that 17. If AiA2A^...A.2n + i is PA^ + PA3+...+ PA,„ ^^ = PA, + PA,+ 18. From any diculars are ... + PA,„. point on the circumference of a drawn polygon of n sides : circle, perpen- the sides of the regular circumscribing shew that to (1) the sum of the squares of the perpendiculars is (2) the sum of the cubes of the perpendiculars is 21 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight —— — ; . 2 339/467 8/12/2019 Elementary Trigonometry by Hall and Knight CHAPTER XXIV. MISCELLANEOUS TRANSFORMATIONS AND IDENTITIES. Symmetrical Expressions. An 301. expression certain of the letters is it said to be syiametrical with respect t<i contains, if the vahie of the expression remains unaltered when any pair of these letters are interchanged. Thus + cos j3 + cos tan (a + tan (^3 - cos a sin a sin -y, (9) <9) +tan (y /3 sin - 6), -y, are expressions which are symmetrical with respect to the letters a, /3, y. A 302. symmetrical expression involving the sum of a number of quantities may be concisely denoted by writing down one of the terms and prefixing the symbol 2. Thus 2 cos a stands for the sum of all the terms of which cos a is the type, 2 sin asin ^ stands for the sum of all the terms of which sin asin /3 is the type ; and so For instance, the three letters 2 cos 2 symmetrical with respect to is ^, y, = cos /3 cos y + cos y cos a + cos a cos /3 (a- 6) = sin {a-d) + sin (/3 + sin (y - 6). /3 sin the expression if a, on. cos y ; (9) A symmetrical expression involving the product of a number of quantities may be denoted by writing down one of Thus 11 sin a stands for the factors and prefixing the symbol 11. the product of all the factors of which sin a is the type. 303. l^'or tlic instance, if the expression three letters a, is synunetrical with respect to ^, y, n tan(a + d) = tan (« + ^) tan (^ + d) tan (y + II (cos /3 + cos y) — (cos j-i + cos y) (cos http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight y + cos (9) a) (cos a + ci>s (i). 340/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight SYMMETRICAL EXPRESSIONS. 297 With the notation just explained, certain theorems in Chap. XII. involving the three angles A, B, C, which are con304. may nected by the relation ^1+5-1-^=180°, For instance concisely. sin 2 A 2 Example sin A = 411 cos — 2 tan A =11 tan 2 tan — tan -^ = = c and b &in d cos (( -e sai ^— . (f> Note. d + cb —^ This result + 5 sin0 = c. =0, c — 0; be sin tp — . . sin — —+ . (b-e —r- ~r— +0 2 sin ^— sin Dividing each denominator by 2 sin COS from the equationK c c a 2 cos - sin ^ - cos cos 1 : +b b - sin ; ; « cos cross multiplication a sin h : .1 ; we have cos d (( and A sin 2 the given equations, whence by = 411 Find the ratios oi a 1. acos6 + From be written more , sin ' sin(rf>-(y) we have 0~d> ^^ important in Analytical Geometry. is remarked that cos(^-0) is a symmetrical function hence the values obtained for of d and 0, for cos [0 - 0) = cos (0 - ^) b c involve 6 and symmetrically. a should It 1)6 ; : : Example 2. If a the equation a cos d and +b /3 = c, sin 4 cos- - cos- ^ are two different values of d which satisfy , find the values of sin a + sin /3, http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight sin a sin /i. 341/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 298 From the given equation, by transposing and squaring, (a cos .-. The [OHAP. (a2 «? + - = c)2 62) sin2 Z>2 cos2 (9 = 62 - 2ac cos ^ + (9 _ cos2 ^) 1 ( - c2 62 roots of this quadi'atic in cos 6 are cos a .•. + cos a cos i3 •^ = -s— and cos p ; (1), ,., a2 + 6- c- - cosacosS=-5 = 0. — 62 ja--\-h- and 4cos2- cos2^ = And +cosa)(l + (l +-ii a- + a' From the data, we By — + p> 6- a- b cos d cos last two results (6 sin • Example If a 3. the equation a cos ^ values of a and jS ^- and +6 c)2 /3 ^ - b^ J, + a- ,., b~ c2-a2 a- + b- also be derived = a2(;os2^ = fl2 = c, (1 from the equation _ sin2 0). prove that tan shew that 1 substituting 26c ,-5 are two different values of 6 which satisfy sin d are equal, — + sm B=- may = we have Similarly, from equation (2) a \ 26i- a- sm becomes (1) ^J , . These - - S + sin/5=-s sin a or fir \2 which c. = + \ fw - - a + V2 J 62 -^; writing a for 6 and 6 for a, equation cos By + - c2 - a and -^- ^ are vahies of sin d a 5:5 0' cos/3) 62 see that „ satisfy the equation — + 2ac =1 (2). + b- i + tan2 ^ http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight Also . if the = c-. - tan2 cosg=1 a- -^ = - 2 tan and sin0 = ^ \ + tan2 1 342/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight SYMMETRICAL EXPRESSIONS. XXIV.] (( ( 1 - tan^ H ) = c, we + b sin 6 + 26 tan - in the given equation a cos =c a that (c is, The , tan a , + a) tan^-- 2 ^ + tan 2ft ^ 2 c+a 2 If the roots of equation (1) a^ Note. The substitution Analytical Geometry. Example 4. If cos 6 + 0) and From sin 2d + cos + sin + c+a) :i sin 6 +a ; a + frequently is sin ((> = b, used in find the values we have + sin d + cos h cos a tan »+ = instead of tan + - c 20. sin 6 cos(. c.-a a)\ employed 2 t = we have = a and ^ For shortness write a b^=c^. here the given equations, .'. ; ; tan - tan^ 2 2 lr=(c + a){c- of cos [d j (1). a^ ^ \ ' are equal, whence ^ = and tan ^ and^ , +a c tan^ tan jj+(f- a) 2h = j3 + a roots of this equation are tan ~ .'. have 1 ( 299 , 0) ,. . (8 <f>) = — \-t, b - ' . a —^^ = then ; a -b ^-„ 2a6 ^t • sm and + 1 + ;2 - a2 + 62 Multiplying the two given equations together, sin 26 .-. + sin 20 + 2 sin2tf sin (^ + sin20^2«6| + 0) 1 - http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight we have = 2rt 6 „ ^, ) 343/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 344/467 8/12/2019 Elementary Trigonometry by Hall and Knight Example In any triangle, shew 7. Sfl^cos J =abc /.•=-. ' Let so that by k'\ — sin =k a (1 tliat + 411 cos j1). ' a By A = 7 sin A^ =k b B sin C sin sin B, c ' = k sin we liave to prove that sin ^ cos Now A = sin A Hli sin ' A cos B sin A = 'i^ C sin A sin'' (1 + 411 cos = 22 has been shewn in Example 2 and sin sin 2^ 1, Ai't. 24 =411 2 .-. 44 sin = 82sinM 2 .. -2 sin 4,4 ; 135, that sin .4; - 4n cos sin* .1 2A= - 32n sin 4 sin 4 =811 sin cos4 =11 4 + 3211 sin 4 (1 + sin Jf d = a, , a cos b X 3. It i'<).sa cos a+/3 ——- = + Tsina = l, n+ y . 7 sin b . ; llcos^, a. , l> . sin c a+/3 -— = - Solve the simultaneous equations - 4 A + y Sin ^=-, (/ ,, cos 11 satisfy the e(|uation and ^=/3 - COS^ prove that . 411 cos 4). EXAMPLES. XXIV. 2. sin 2.4 easy to prove that it is 1. A). sin 2 = 22(1 -cos 2.-1) it C. substituting these vahaes in the given identity, and dividing ^2 and 301 SYMMETRICAL EXPRESSIONS. XXIV.] , «=1, c cos a-/3 —^~ : - cosj3+^siu/3 = y -sin « - 7-cosa .V a . . b http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight = l. , l. 345/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 302 If a and ^ are two [OHAP. a cos d + different solutions of h sin = c, prove that 4. cos(a+;3)= ,o .. , cos'' 5. . = —^r— -g^—to . (a^ -J- 62)2 7. .s.nâ 8. If + / • 9. . , a cos a + 6 siu(a If , COS 6 cos ^ _ 2c2 (a^ - /y^^ . = a cos j3 + h 2ai ^x fe2) ^^,^_^^,^/ sin a + /3) = -s cos^4-cos0 , rr,, = a and ^ and sin ^ sin COta (3 = c, prove 2a6 ^ + COt8 = -5 , + sin ^ = 6, tliat ;,. prove that + J2)2_4j2 4(a2 + 62) (a2 = ^. (a2_52)(a2 + J2_2)'. + cos2(^=^ ^-^ 10. cos2<? 11. tan 6 12. tan - + tan 13. Express + tan 2^ 1 2a2 (a2 + sm^/3 = -- - Sab (/> = ^^2-^2^2^:4^ • 2~«2 + 52 + 2a- cos- a - COs2 ^ — C0S2 -y 4- 2 COS a COS /3 COS y product of four sines, and shew that it vanishes if any one of the four angles a±/3±y is zero or an even multiple of n. as tlie 14. Express sin2 a as tlie 15. + sin2 /3 — sin^ y + 2 /3 cos y product of two sines and two cosines. Kxpress sin2 «-f sin2/3 as sin a sin ilie ))ro(luct + sin2y- 2 sin n sin /tJsin y- 1 of four cosines. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 346/467 8/12/2019 Elementary Trigonometry by Hall and Knight 16. ' prove 17. prove 18. 303 AI.TERNATINC4 EXPRESSIONS. XXIV.] COS^ If = a - COS a COS /3 :; 1 th;t.t one value of tan - If tan2 tliat tan^ - (9 = sin a j3) . prove that one value of tan- is siu . /3, tan - tan - is tan ^ (cos a + sin If y tan - cot - is cos2^^^—-^ one value of — COS 8 COS a . = sin a cos /S, a.tan fn tan — fi^ -^ I In any triangle, shew that B sin C= 2abc (1 + cos A cos B cos C). 19. 2a3 sin 20. 2acos3^ 21. 2a^cos(5-(7) = ^(l-4cos4cos5cosC). = 3afec. and ^ are roots of the equation « cos ^ + 6 fomi the equations whose roots are If a 22. and sin a (1) sin ^ cos 2a (2) ; and cos sin (9 = c, 2/3. Alternating Expressions. said to be alternating with respect to 305 An expression sign of the expression but certain of the letters it contains, if the pair of these letters not its numerical value is altered when any is are interchanged. Thus cos cos2 a sin a- cos ^, (/3 tan(a-/3), sin(a-^), - y) + cos2 /3 sin (y - a) + cos2 y sin (a - ^) are alternating expressions. Alternating expressions 306. the symbols 2 sin2 „ 11 „i,i tan 2 and 11. Thus O - y) = sin2 a sin (/3 - y) = tan {fi - (/3 - may v) be abridged by means of + sin^ ^ y) tan (y sin (y - a) + sin2y sin(a-/3); - a) http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight tan (a - ji). 347/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 304 We shall confine our attention chiefly to alternating expres- and we shall adopt the y-a, a — ^ in which /3 follows a, sions involving the three letters arrangement cyclical y follows cos (a + d) ^-y, and a follows /3, Example i; [cHAP. sin - 7) (/3 ^, y, y. 2 Prove that 1. a, cos (a = S (cos a = cos 6 2 + ^) sin (/3- 7) = 0. cos 6 - sin a sin 6) sin cos a sin - (/3 y) - 7) - sin H 2 sin a sin (j8 (/3 - 7) = 0, S since • Example sin 2 cos a sin Shew 2. - 7) (/3 + (1) = - 7) S that sin 2 (7 - a) and 2 sin 2 sin (^ - a) cos (o =2 sin(a-/3) {cos =4 sin (a - - 411 sin /3) (/3 + sin 2 =2 = Example (jS sin a sin - 7) (a +^ - (a -/3) - 4n - 7) sin = 0. (/3 - 7). - ^) 27) +2 -cos sin (a - 7) sin (j3 sin (a - j3) cos (a -/3) (a-l-/3-27);• - 7) (/3- 7). Prove that 3. (1) 2tan(/3-7)=ntan(/3-7); (2) 2 tan From = (^3 ji tan 7 tan (^ - 7) Art. 118, + tan A it A + B+ tan /.' - IT tan C = 0, we (/3 - 7). see that + tan C = tan A tan B A=§-y, Hence by writing = li tan C. = y-a, C=:a-p, we have Stan(/3-7) = ntan(/3-7). (2) From the IbrmuliB for ta,n(j3-y), tan (7 -a), tan(a-/i), we hav<> - 2 (1 + tan ;3 tan 7) tan (|3 - 7) = 2 (tan /3 tan 7) =0; wlicnce by transposition 2 tan [i tan 7 tan {ft - 7) - 2 tan = -n tan - http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight (j8 (/J - 7) -7). 348/467 8/12/2019 Elementary Trigonometry by Hall and Knight SYMMETRICAL AND ALTERNATING XXIV.] Example Shew 4. S cos 3a that sin =4 - 7) (jS 2 cos 3a sin (^ - 7) Siuce cos 3a sin — sin - 7) (/3 (3a - sin cos (a + /3 - + /3 + 7) + (3 - a) IT sin(/3-7). 7) - sin (3a - 7) - sin (3a -7+ (3)3 (3a = sin we have 22 - /J + 7) (37 + a - + sin + sin j8) + y), (i cos 3a sin (/3 a) - sin (87 - a + /3). fifth terms, - 7) = cos(a+/3 + 7) = 4cos The 307, +7 - (3/3 Combining the second and third terms, the fourth and the sixth and first terms, and dividing by 2, we have S 305 EXI'llKSSIONS. [sin 2 + /3 + 7) (a (/3-a)+sin2 (7-^8) + [See II sin (/3- 7). following example sin 2 (a given is a.s -7)} Example 2.] a spcciuien of a concise solution. Example. tan .1- and If .f (1 and .r If wc +y - tan /3 tan sin 2a (tan /i 7) +y + v/ find the values of a) (tan a = (tan ^ - tan 7) + tan- a — (1 + tan- a) (tan /3 = sec- a (tan /3 _ sec a sin {^ - + + i/) tan 7 - 0, {x + z tan a tan ^ — x + y + z, 2/3 + ^ sin 27 = 0. sin we have - tan 7 tan (1 + tan 7) + y denominator of x = (1 - tan 7 tan ji tan 7 tan a the given equations, .(• x) tan ) tan y j8 prove that From + z tan a+{z + (// (tan .r : 7 + tan y + tan /i) (tan jS a) : - +^ a) ^• (1 ^ - tan a tan + (tan a tan /i) — 0, /3) — 0. z by cross multiplication, the (1 - tan a tan ji) (tan 7 + tan a) - tan 7) - tan 7) tan 7) 7) ' cos a cos /3 cos sec o sin .-. X sin 2a +y 7 (/3 - 7) sin 2/3 +3 sec j8 sin (7 sin 27 = - /cS sin — '2k1, sec7sm(o-/3) a) 2a sec o sin sin a sin (/3 - (/3 - 7) 7) -0. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 349/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 306 [cHAP. Allied formulae in Algebra and Trigonometry. some lu the identity 1. -a)(b- {x put a; then h .-. Example + {x-b)(c-a) + c) = 008 2^, .T and ((=cos2a, t -c = cos 2/3 - 2 sin (a + ^) cos 27 = then sin^jS 6 sin-'a sin + 7) sin (;8 - 7) a (coSjS- cos a, 7)= - 2 b - 7) — sin-7; sin = - + 7) (^ II sin b (/3 - 7) + 7) . 11 sin II (cos (fi - 7). + c)\\[b-e), — cos^, (cos a cos a (cos /3 c = coS7; + cos /3 + cos - cos 7) = 7) ^- cos 7)* ; a - 3 cos a) (cos /i - cos 7) (I cos'' = -4 (cos 2 (/3 sin In the identity 3. But 2 sin(a-^), O) c), - sin'*7= sin a = cos .'. = co8 27; + 7) sin(^-7)==0. c (j8 (/J put cos'' (/3 f + 7) ~n{b- :â^{b~c)^ -{a + 2 (^8 = sin-;8, b-cExample + [a - 2 sin sin (a - S) sin a=;sin-a, 2 = cos2;8, 0, In the identity 2. put .-. -c){a-b):^ {x - a =: cos 2^ -cos 2a = 2 sin 2a2 (b-c)= that deiUicc interesting trigonometrical identities. Example .'. we can P'rom well-known algebraical identities 308. a + cosy3 + cos7) II (cos /j- cos 7); is, cos 3a (cos Example Here is ^ li a-\-b 4. a, b, c condition - cos 7) /3 may - 4 (cos a + cos /i + cos 7) + c — 0, then 11 we put a=:cos p- cos 7). + b^ + c^ — Sabc. d^ be any three quantities whose satisfied if (cos (0 + sum ^) sin (^3-7), is zero; this and b and c equal to corresponding quantities. Thus 2 co« (a ' 1 0) sin'' (/i - 7) = 311 cos http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight (a + 0) sin (/i --<,). 350/467 , . 8/12/2019 Elementary Trigonometry by Hall and Knight ALGKBRAICAL IDENTITY PKOVED BY TRIGONOMETHY. XXIV.] An 309. may sometimes algebraical identity 307 be establishecl by the aid of Ti-igonometry. Example .<• By (1 - U . (1 //-) putting a; + y + z = xyz prove that - --) + (1 - ^-) (1 -x ) + z(i.v , = taua, tan a = tan/:J, // + tan tan a whence == - tan + tan 7 j--^L__L= /3 a a '2a .-. tan 2a + 2x ... .^ (1 _ it is _ ,0 , \ , where n an integer; is + j3 + 7 = n7r; + tan 27 = tan 2a tan ^2 +y have easy to shew that tan 2^ ,2) = i-vy^- + 2p + 2y = 2mr. 2t/ y2) (1 >/} - tan (^ + 7) = mr-{^ + y), .-. this relation = tan 7, we 2 - (1 + tan 7 = tan a tan ^ tan 7 /3 .• From .1-2) .V (1 - _ ^2) +2 (1 _ XXIV. EXAMPLES. Prove the following identities tan 27 8xyz _ .2) (1 2j3 .^2) (1 _ ,/) ^ ixyz. b. : 1. 2sin(a-^)sin(/3-7)=0. 2. 2 3. 2sinO-7)cos(^ + 7 + (?)=0. 4. 2cos2(/3-7) = 4ncosii3-y)-l. 5. 2 6. 2cot(a-^)cot(a-y) + l=0. 7. 2sin3asinO-y) = 4sin(a+/3 + y)nsinO-y;. 8. 9. 10. cos sin /3 /iJ cos y sin sin (/3 y sin - y) = :S sin - 7) = - /ii H sin sin {(3 7 sin (/:i - 7). - y). 2cos3asinO-y) = cos(a + ^+y)nsin(/3-y). 2 cos ((9 + a) cos + y) sin (^ - a) sin (^ - y) = 0. 2 sin^ /3 sin- y sin O + y) sin O - y) = - n sin http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight {(3 + y) II «in l/:* - y> 351/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 308 Prove the following identities 11. 2 cos 2j3 cos 2-/ sin : O + y) sin (^ — y) = - 411 12. 13. 2 2 cos 4n sin (/3 sin 3a (sin /3 + y) sin 03 - y) =— [cftAP. sin (/3+y) 811 sin (/3 11 sin ifi-y). . + y) n . sin (/3 - y). sin y) = 4 (sin a + sin /3 + sin y) IT (sin /3 - sin O + y) - = SH sin 14. 2 15. 2cos3(/3 + y + ^)sin3(/3-y) sur'' sin'' ()3 y) = 311 16. If x+y + z~xyz^ ^ 17. (^ cos + y) H (^y sin (^ . +y+ (9) - y). n sin . y). (i3 - y). prove that 3^--.t-3 3.y-A-3 i-3A-2~ 1-3a-^' If y^+â;+,«;y=l, prove that %jc{\-y^){\-z'^) 310. From a = Axyz. trigonometrical identity many others ma\' be derived by various substitutions. For instance, if ^1, ii, c;onnected by the relation sm 1, , ^1 + sni zj + sni J==7r-2a, Let thei C are any angles, positive or A+B + C=tt, we know that sin i Also 2{a A — sin ,1 C , B = 7r-2^, sin 2a ..,, Agam,let ., then B — C cos - . A — = sin a. + ^ + y) = 3n-{A+B + a) = 2n; a+i3+y = + sin 2/3 ,7ra 7r, + sin ,, 2y = 4 sin a sin TT , sni yl = cos a • 2' -.1 and cos — , , http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight /3 sin y. /(Try ^ 2-2' 8 ^1 = 2-2' ^ = 2-f' . cos C'=7r-2y; and cos 2n, .-. and Î = 4 cos — negative, 2 — cos n—i. -,— 4 . 352/467 . 8/12/2019 . Elementary Trigonometry by Hall and Knight Also + ^ + y = 37r-2{A+B+C) = ^n-27r; a a J and cosjr 2 y = 4cos TT-a ^ + cos^ —- cos tt-v -^ + cos^ — cos 7r-/3 —-' Example. cos . A , 2 A —+ 4 2 ^+Z? + C = 7r, shew If B — COS 2 - COS 2 ^, ^ ^ ir +A C ^ = 4 COS TT 2 COS ^ = cos a - ( that +A TT —— —+ B COS TT-C TT COS 4 - 4 4 C- 3 1~ 2' = sin a, and ) ^ 4 4 ir+B a 4^=2' then 309 IDENTITIES DKIUVED BY SUBSTITUTTON. XXIV.] y ir ir cos 2' — — cos ( 7 + ^ ) — - sin 7, so that the above identity becomes ^ • sni a which is clearly + sin /3 + a sin yy hen A tan When n — 0, (. ( IT 2 ^ = 11 tan is sum If tan a ; C^ tan = - tan C ; ^4. is satisfied in B = y + a-2^, the case of any if C=a+/3-2y. O + y - 2a) = 11 tan (^ + 7 - 2a). cot (7 + a - /3 Put + 7-a = ^, then, by addition, /3) that cot (a .-. cot whence Scot (7 + /3 - 7+a-/3 = B, J+£ + C = a + H. K. IT as for instance + /3 + 7 = 0, shew S is, - (« n- the given condition A=^+y-2a, that , ^ =2+2=''A+B+C +B) = tan ^ three angles whose Example. ^ +B+C=nn, whence we obtain Hence V true since a+p + y=-TT + 311. /3 . 7 = 4 cos - cos - cos |3 a 7) =1 + ^-y=C; + 7 = 0; (^+B)= -cotO; 2 cot + a-/3) cot A cot B = 1, (a + /3-7) = l. E. T. 22 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 353/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 310 The 312. manner in following example which an identity Example. identity. Prove that 2n In Example cos 5, + 7) + n cos 2a = S cos 2a cos- (/3 + 7) (/3 we have proved that Art. 133, 4 cos a cosj3cos7 = Scos(/3 In this identity tively, replace first and secondly replace Thus a further illustration of the be established by appropriate is may some simpler substitutions in [cHAP. 8n j3, 7 by a, ^, 7 by {/3 4n cos 2a + 7, 7 + a, a+/3 /3 2a, 2^, + 7) = 22: cos 2a + cos and a, + 7-a) +cos {a+^ + y). 27 2 cos 2 (a = - cos 2{j3 + y - +^ + 7), + cos 2 (a + ;8 + 7) +3 cos 2 (a + /3 + 7) a] respec- respectively. whence by addition 811 cos (/3 + 7) + 4n cos 2a = 22 cos 2a + S cos 2 = 2S (/3 +7- a) cos2a + S {cos 2(/3-f-7-a)+cos2(a + /3 + 7)} = 22 cos 2a + 22 cos 2 (|3 + 7) cos = 22 cos 2a {1 + cos 2 (/3 + 7)} = 42 cos 2a cos- (/3 + 7) 211 cos .-. (/3 + 7) + n cos 2a = 2 cos 2a 2a cos* 03 + 7). Suppose that A'B'C is the pedal triangle of ABC, and let Z>', c', the sides and angles of the pedal triangle be denoted by and A', B', C, and its circum-radius by R'. Then from Arts. 224 313. <?-', and 225, we have a'=acosJ, A' = h' B' = \m°-2A, By means =h cosB, \m°-2B, of these relations, for the triangle ABC In the triangle 2a c' = ccosC, 11' =— , = 180°-2r. C we may from any identity proved derive another, as in the following case. ABC, we know cos A=^4tR sin that A sin ^ sin C hence in the pedal triangle A'B'C, 2a' cos.4' = 4/iL'sin J.'siu /?'sin .-. that is, Sacos^ cos (180° - 2a cos A cos - 2A) = 2im 2A — 2R sin http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight sin 2A C; (180°- 2J) sin 2/i sin 2C. 354/467 . 8/12/2019 . Elementary Trigonometry by Hall and Knight IDENTITIES DERIVED BY SUBSTITUTION. XXIV.] In any triangle Example. a^ cos^ ABC, shew B-c cos C A-h ^ cos26c cos i? cos 311 that C COS 2^. In the pedal triangle A'B'C, we have 6'2 + c'2-a'2 = cos^'; -—-, hence, by substituting the equivalents of b^ coB^ B—+ c^ cos C - J cos^ A 2bc cos B cos C V, a', c', A', we have ,,„^„ ,^ ,, 180''-'2.4)- =:cos -cos2^ .-, ; whence the required identity follows at once. be the ex-central triangle of ABC, we may, as in the preceding article, from any identity proved for the triangle ABC derive another by means of the relations If 314. a^ A ^B^Ci = a cosec — 6^ , = 6 cosec — The following Exercise 315. Cj , = o cosec — , î = 2i?, consists of miscellaneous questions on the subject of this Chapter. EXAMPLES. 1. Shew that 2 2. XXIV. c. , cot (2a + /3 - 3y) cot (2i3 4- y- 3a) =1 Shew that 211 sin (/3-f y) (1) (2) + 11 sin 2a =2 sin 2asin2(|3-t-y); n sin {^ + y-a) + 2U sin a = 2 sin'- a sin (^ +y- a). In any triangle, prove that 3. (1) a2 cos2 A - 62 cos^B = Rc cos Csin 2{B-A); D A (2) a^ cosec^ (3) 2 (6 cos k j5 ~ -i ^^ cosec^ —= c cos (7) cot 4flc cosec J=- OH—— — sin 4 ^ ; 2i?2 cos 2 .1 22 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight — 355/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 312 sin 2$ If 4. =2 cos 26 = cos 2a cos and prove that one value of tan 6 sec a cos ^ prove that sin^ 2S, 8. y sin If y = (sec a - cos ^ — cos a _ cos ^ - cos /3 ^ (^ = cot a tan y is 1) (sec /3 - sin- a cos sin^ /3 1). /3 cos a gtan- tan and tan 6 = cos n tan one value of cos 6 pro\'e that is cos j3, sec y. /3 are two diflferent values of 6 which satisfy and If a = sec /3 cos ^ = cosy, one value of tan ^ ]ii'i)VP tlijit 8. 2y cos tan ^ tan 6.4) and 7. = cos 2j3 tan-tan^ = tan^, If 5. is sin a sin y, ^ be cos 6 cos (f) + ac sin ^ sin = ab, (f) ])rove that (b- 9. + c- - a-) cos a cos /S + (c- + a- - 6^) sin If /i , cos ^ that prove ^ 10. 1 f /3 prove that 11. If + cos a sin ^ = cos a 3 cos y — 3 cosâ and y are two sin p-H j3 r— sin = a- + /•- cos and y are two satisfy a, j3 + 1=0, + sin y) + 1 = 0. different values of 6 (/> c-. which satisfy different values of 6 cos y + ^ (sin - sinâ cos a cos 6 + k (sin a + sin 6) /3 sin Z>2 _ y ^=1. k ^ cos ^ cos prove that /3 and y are two different values of 6 whicli sin a cos 6 , a sin which satisfy sin ^ sin cos^cos_y_^ sin^sin http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight _y^^^^_ 356/467 8/12/2019 Elementary Trigonometry by Hall and Knight CHAPTER XXV. MISCELLANEOUS THEOREMS AND EXAMPLES. Maxima and Minima. Inequalities. 316. are in The methods of proving many cases identical with trigonometrical inequalities by which algebraical those inequalities are established. Example We Shew that a^t&n-$ + b^cot-9>2ab. 1. have a^ tan^tf + = {atane-b cot df + 2ab b^ cot^ e .-. + b^ cot^ e>2ab, S = 0, or at&n^d = l. a^ tan2 e a tan^- 6cot unless ; In this case the inequality becomes an equaUty. This proposition minimum Example (1 Shew 2. 1 Since may value of a'^t&n'^d + sin^ - sin o be otherwise expressed by saying that the + b^cot'^d that + sin^ ^ > and + sin ^+ + 1 + sin^/S > 2 sin j3, Adding and dividing by + \Vhen 2, + is sin- a + sin* ^ > sin2a 1 >2 1 sin* a 3. sin a sin a sin /3. a)^ is positive, similarly Example 2ab. is sin a ; 2 sin a sin j8. we have sin2/3>sina + sin;8 12 sin ^ - 9 sin^ 8 a sin o sin/3. + maximum The expression = 4 — (2 - 3 sin ^)^, and is therefore when 2-3 sin ^ = 0, so that its maximum value is 4. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight ? a maximum 357/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENT ART TRIGONOMETRY. 314 To Jitui 317. the nwmerwally greatest values of a cos ^ + b sin 6. a = rcosa and 6 = rsina, Let and tan r'^=ia?'-\. 62 so that a cos ^ + 6 then = r (cos sin ^ = rcos Thus the expression is is, a a numerically greatest wlien = ±1 the greatest positive value = ?• = Hence, + 6^, value = — r= - v« + sja?- value of a cos ^ + 6 sin ^ + and the minimum value is c The expression a = (a c is c + \/«^ +&^ - v «^ + ^^• cos (a + ^) + 6 cos cos a + 6 cos ^) cos ^ - O + ^) (a sin a + 6 sin /3) sin ^ and therefore its numerically greatest values are equal positive and negative square roots of (a cos a that 6 ^. c2>a2 + fe2^ if maximum 318. ; — a). and the numerically greatest negative the =- 6 cos a + sin ^ sin a) (6 cos (^ - a) that [CHAP. + 6 cos /i)^ + (a sin a +6 sin to the /j)^ are equal to is, sjd^ + + - 62 + 2a6 cos (a /3). In like manner, we may find the maximum and minimum values of the sum of any number of expressions of the form a cos (a + 319. ff) If a and whose sum of cos a or a sin (a is /3 + ^). are two angles, each lying between given, to find the value of cos a cos /3 and + cos ^. Suppose that tlicu maximum and - 2 cos a cos a+/3=o-; /3 = cos (a + /:<) + cos (a-/3) — cos + cos (« - (i), (T http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 358/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight MAXIMA AND MINIMA. XXV.] and is therefore a laaximmn when Thus the maximum vahie Again, cos a a — /3 = 0, of cos a cos is therefore a maximvuu when Thus the maximum vahie or a /3 is o = /3= cos^ —^ cos — + cos /3 = 2 cos = 2 cos and 315 cos - . _^ . g- , = /3=^. of cos a + cos ^ is 2 cos ^ . Similar theorems hold in case of the sine. Example maximum If 1. ^, C -B, are the angles of a triangle, find the value of sin A+sinB + sin C and C . sm A + • sin This expression . i> is B sin ^+B —- =2 cos — cos — ^ . sin C. A and B — cos A-B — + sm C — 2 a sin remains constant, while Let us suppose that • A of sin . 1- sin vary. ^ t-sin C. maximum when A=B, Hence, so long as any two of the angles A, B, C are unequal, the expression sin .4 + sin iJ + sin C is not a maximum; that is, the expression is a maximum when A =B = C = &0°. Thus the maximum value = 3 3 sin /3 60°=-—- Again, 2 sin A sin B sin C= = This expression is a {cos (A-B)- {cos (4 - (A+B)} sin G + cos C}sinC. maximum when A=B. Hence, by reasoning as before, sin value jB) cos A sin B sin C has its maximum when J = ii = G = 60°. ~ Thus the muxiuiuiu value = sin*60°=— http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight . 359/467 .. 8/12/2019 Elementary Trigonometry by Hall and Knight KLEMENTART TRIGONOMETKY. 31 1) Example — whose sum , We have and If a 2. /3 sec o + sec a ^ 4 cos cos (a ^ = cos 1 = cos a Since a + ;8 is ^ , u-hose sum a-jS 2 cos ^-t^ value of sec a — 2 cos —2^ gd-^ —— - cos^ +B + sm -y^ cos is, a=^— when 2 sec is are n • a-S ' /3 a + /3 + 'y Let . ^- cos— ——- sin + /3 -2- a when the denomi- . +B —- a — angles, each lying between constant, to find the cos a cos —2— oa + - sm^ ——/ a . + sec/3. cos a + cos ^ -5cos a cos /3 — constant, this expression is least a, 13, y, 8, is coS|8 \ Thus the minimum value If 1 1 + cos(a-S) + j8) r/ 1^' nators are greatest; that 320. minimum —^ cos —~^ * and are two angles, each lying between constant, find the is [CHAP. maximum a7id value of cos y cos h = +S+ <t. Suppose that any two of the angles, say a and ^, are unequal then if in the given product we replace the two unequal factors cos a and cos /3 — and by the two equal factors cos -— cos —^ the value of the product is increased while the sum of the angles Hence so long as any two of the angles remains unaltered. a, i3, y, 5, ... are unequal the product is not a maximum; that In is, the product is a maximum when all the angles are equal. this case each angle ° =- n . Thus the maximum value In like the is cos n manner we may shew that maximum value of cos a + cos ^ + cos y + http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight =n cos - . 360/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight MAXIMA AM> MIMMA. XXV.] The methods of 321. 317 examples sohitioii u.sed in the Ibllowiug are worthy of notice. E.vampli' ,,, We Shew 1. . , have that tan 3a cot a cannot he between 3 and - tan 3a = I . tan 3o cot a tan a tan2 a .: — a and If 2. greater, find the 3n - then tan'^ less b-t^ than x^ and which a > t (b- - a-) {x^ x->a^- ^ =f t-); - may be real, a-) b^. ^Ja^ - Example k are b, c, b^. constant quantities and a, + b tan j8 + c tan variable quantities subject to the relation a tan a By (a'^ value of tan- a + tan- /3 + tan- /9, y y=k, 7. multiplying out and re-arranging the terms, we have +b^ + C-) = {b tan (tan- a 7 - c -f tan tan ^)- -l- /3 4- (c But the minimum value hence the minimum (a- that the -b tan 0. found from this equation is minimum is + 2bxt + x -a^ = 0. Thus the minimum value find the there- 0>aâ^-b^-x-); .: a, positive, . + 2bxt + x^ = a^{l + t-{b'^-a^) :. If . 37i must be a and put tan .r, In order that the values of 3. 1 — ajl + t'^~bt; x b —- 1 value of a sec Denote the expression by .: 5— = n say a h are positive quantities, of minimum .-. : n—B d—n = =~ ^ n must be greater than 3 or Example 3 - tan- a - tan^^ These two fractional values of fore — 1-3— — . is, the + tan- 7) - (a tan a tan a- a tan 7)- -f + (o b tan ^ tan +c ^-b tan 7)- tan of the right side of this equation o)^. is zero; value of t--i-c2)(tan-a-t-tan2^ minimum + tan2 7) -k'^ = ; value of tan- a -t- tan- 8+ tan- 7= ' -^ a^ — ——„ ^, + b^ + c- http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 361/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 318 EXAMPLES. variable find the is XXV. mininmm [CHAP. a. value of the followiug Wheu 6 expressions jo I. cot ^ + 8sec2(9 3. '^ tan 0. + 18cos2^. 2. 4 sin^ 4. 3 - + cosec^ 0. 2 cos ^ + cos^ (9. Prove the following inequalities tan2a+tan2|8 + tan2'y>tan/3tauy 5. + tan y tan a + tan a tan sin2a + sin2/3>2 (sin a 6. When is variable, find 7. sin ^ + cos ^. 9. acos (a If (r =a+ and - , and where is o- /3- sin a cos p 10. ^. and /3 1). maximum the 8. + ^) + 6 /3, + sin j3. value of + J3 sin 0. cos ^ + 5- sin (a + ^). are two angles each lying between constant, find the maximum minimum or value of sin a 13. tan a + tan C If A, B, or + sin II. minimum 15. 7. cos /3. ABC cos B cos C. — + sin^ — + sin2— 2 2 tan'''-+tan2-- 20. cot2 If sin a sin 14. cosec a + cosec are the angles of a triangle, 19. 21. 12. /3. p}. maximum find the value of ^1 sin ^ /3. A + tan2 — + cot2 B + cof^ b'^<iac, find the a sin- . 2 C. 16. cot A + cot B + cot 18. sec ^ + sec ^+ sec (7. C-'s«2tan . [ Fsc 2 cot — tan — =1. B cot C= 1 maximum and minimum + b sin cos http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight C. .] values of + c cos- 0. 362/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight XXV.] 22. ELIMINATION. If /3, (1, y lie sina 23. greater, and If a 319 and -, shew that between + sin/3 + sin'y>siu (a+^+y). h are two positive quantities of which a shew that a cosec 6>h cot Q + \j or - 24. Shew that 25. r incl the 26. If a, , —^ —+ tan^ r maximum 6, c, ^ ^r sec2^ the 6^. Ues between 3 and 3 value or , — tan2^-cot-^+l — 5-; ; ttt—; tan'' ^4- cot- ^ 6-\ are constant positive quantities, X- is . and «, /ii, y variable quantities subject to the relation a cos a + 6 cos /3+ find the minimum cos â c cos -y = h, value of + cos'-^+cos- y and of a cos- a + 6 cos^ /3 + c cos- y. Elimination. No 322. general rules can be given for the elimination of some assigned quantity or quantities from two or more trigonoThe form of the equations will often suggest metrical equations. special methods, and in addition to the usual algebraical artifices we have at our disposal the identical relations subsisting between the trigonometrical fimctions. Thus suppose it is required to eliminate 6 from the equations shall always A'cos^=a, h. sec^=-, and tan^ = T; a Here but for yco\.6 — all values of 0, we have sec^^ — tan-^=l. .•.by substitution, \. From example we sec that since 6 satisfies two equations which is sufficient to determine 6) there is a relation, independent of ^, which subsists between tlie coefficients and this (either of http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 363/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 320 constants of the eliminate d, To equations. and the determine [CHAP. relation this we result is called the eliminant of the given equations. following 323. Example Eliminate 1. Icosd From and ^^cos _ sin „ tf mr-nq — Iq _ useful + qsm.d + r = Q. 1 -, and , — mp , d vp~ h ~ Iq- mp mr — nq cos some we have by cross multiplication the given equations, cos illustrate between the equations 6* + m sin^ + n = .-. will examples The methods of elimination. „ . sin ' np-lr — ,^^ Iq whence by squaring, adding, and clearing of : - mp fractions, (mr - nq)^+{np - lr)'^=(lq - we obtain mp)'-. p— -m particular instance in which q — l and is of frequent occurrence in Analytical Geometry. In this case the eliminant may The be written down at once and ; we have for I cos +m I sin -vi cos sin 0= -n, 0= -r whence by squaring and adding, we obtain P + m^:=n^ + r^. Example 2. r^ = c- ^ cos From between the equations Eliminate sin the second equation, sin _ _ cos and (tanS = m. we have Jbivl^ + cos^ Jm^ + P' Jm^ + P I in ^ sm g= — „ and cos^ = -, , \Jm^ + p' By substituting in the first ax Jm^ + P equation, by ^ _ we obtain c^ T~m~ ^â^^Tji http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight • 364/467 , 8/12/2019 Elementary Trigonometry by Hall and Knight Example between the equations Eliminate 3. X — cot From 32 KLIMINATION. XXV. J + tan and y = sec 6 - cos 0. the given equations, we have X= 1 „ ^ + tan , tan — + tan2(9 — l tan _ sec2 tan 0' jind V - sec2 1 = sec sec ^ 1 sec ^ _tan''*^ ~ From sec these vahies of x x''-y ' <? and y we obtain = s&(?0 and Bnt =1 ~ tan- ^ sec- .-. .ri/-=rtan^<J. {xhjf-{xi(^)i = \: -.tS?/« that .r*?/^ is, Example EHminate ^ from the equations 4. - a From 1. = = cos ^ + cos 20 and ?^ = sin 4 sin 2^. /( we have the given equations, X , - = 2 cos 361 — - cos - a 2 y , 2 B0 „ f = 2 sm-r-cos- and . 2 2 b whence by squaring and adding, we obtain x But y , - = 2cosp; a 2 = 2 cos2.r /.r2 4 cos* - - 3 cos ^ cos*--3co8 I ( - „0 ( 4 cos- Q ~ i/-\ (x- y- http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight '^ ) ., j ; \ 365/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 322 The 324. following examples [cHAP. instances of the elimination a,re of two quantities. Eliminate d and h»m asin'd + hco&^d — m, <f> From the from the equations (p 1. Example + acos-fji = n, + h cos'^ d = m sin* d .•. (a- m) , „ <p + a cos- <p—n tan ^. d) ; ; b . — — b-n ' n From (sin^ (p + cos^ <p) ' f tan-d)=; .•. . the third equation, =b a- tan- d a^ tan- -b) _b {in .•. .•. a- (bm ?Hai (a -b^- mn + -b)+ nab .•. ?Hfl/v .rcos + 2/sin 5=xcos the data, we see that d .T .: (x cos a (.r- .•. + ?/ cos a - 2a)- = y ' {a- (a - V^) + am) ; + b); from the equations 2sm^sm- = l. . = 2a, sin</) . are the roots of the equation sin a = 2a sin^ a = y- ; (1- cos- a) + y^) cos- a - iax cos a + 4a- - y^ = Q, a quadratic in cos a with roots cos 6 and cos 1 ; b a + J. and +y a) -n 1111 +m -= ; (an - a- - r»« -b) = mn Eliminate 6 and 2. tf =b + nab = mn n .-. Example (a b iw) - {n a-m But ?> the second equation, we have b sin- is b) cos- 6 ~ m (1 which - — 7)1 tan-(?= + cos- (sin^ 6 = (m sin^ .-. From — equation, Ave have first a From « tan ^ =4 sin- -^ sin 1= (1 - cos http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight (9) (1 ((>. - cos(^) 366/467 8/12/2019 Elementary Trigonometry by Hall and Knight 323 ELIMINATION. XXV.] whence + cos cos d <j> âx x^ + = cos d cos ^ _ x'^ i/ -^ + y- y-=ia (a- .'. The method exhibited 325. â^ - y x). in the following example one is frequently used in Analytical Geometry. Example. If a, b, c are unequal, find the relations between the coefficients, when tliat hold acos ^ + i sin^ = c, and (/ cos ^ + 2rt cos ^ sin fl + 6 sin- ^ = c. ehminating 6 from the given equations. This is most conveniently done by making each equation homogeneous in sin 6 and cos 6. The required relation will be obtained From the we have equation, first rt cos 6 + bj' b sin d — c ^cos- d + sin- 1) whence, by squaring and transposing, (a2 From - c ) cos2 d + lab cos the second equation, ^ sin ^ -(- (^^ (a-c)cos-(9 From (1) and (2) c ) sin- ^ = (1). we have a cos- 6 -\- 2a cos ^ sin ^ + b sin- d = c .'. - (cos- d -^ sin- 0) + 2acose sin^-f(t-c)ain20-O ; (2). we have by cross-multipUcation, cos ^ sin ^ cos- d 2ah [h-c)- 2a {b-^ - c^) {b^ - c^) (a-c)- {a^ - c^) {b - c) sin^^ ^2a{a^-c^}-2ab{a-c)' cos 6 sin 6 cos- d sin^ e 2a{a-c){a + c-b)' (b-c){a-c)(b-a) -c)(a- c) {a + c-b) = (b c)^ (a - c)- (b - a)-. -2ac{b-c) .: By by (h - 4ah- {b supposition, the quantities a, -c)(a- c), b, c are unequal ; hence dividing we obtain 4a-c (a + c- b) + (b- c) (a - c) (a http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight - b) = 0. 367/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 324 EXAMPLES. XXV. [chap. b. Eliminate 6 between the equations cos^+f sin^=l, 2. a sec ^ — 3. cos^ + siu^ = «, 4. .*' 5. a .r tan ^ = sin ^ + cos = cot 6 = = cot .V 7. cosec Q 8. 4^7 9. .r — sin = 3« cos 6 y = a^, Tf rt, .r 14. If sin 5 tliat = sin ^ + cos If cos — cos B 6. = ly\ 3a. si n ^ - a sin 3^. y = 6sin ^ (4cos-^- 1\ sin ((9 - yS) = i, + cos ^= a, must hold between ^= - /3). and v .» y=cos ^+sin ^sin and sin 2^ + cos 2^ = {a^-h~lf=a^ ^- sin (a if 2^. />, ('2-a^. and cos 3^ + sin 3^ = a, 6, a = 36 - 26^. Eliminate 6 from the equations ff 17. — sin : y=sec2^ (y-a sec^l 2^, shew that 16. S. .r-y = 4sin2d ^sin shew that 15. 6 4y = 1), and = 3-cos4^, 13. cos - 2ab sin (a - j3) + 6'^ = cos^ Find the relation .r+_y - sec 6 11. 12. 6, — cosec ^=acos ^(2cos 2^- cfi .r. each of the followini^ cases in 10. shew that <? = cot h 0, = = \. y = tan^ + cotrf. ^, tan ^-.r), cos(5-a) cos d ^+y tan = sec 0-k-a cos 3 6, = tan2^(a. b cos 2^ = + cos 6, + tan Z> 1 /, Find the eliminant 6. ^-T -sin a If cos ^ .V - =a shew that 6 sin ^ cos ^ + a- = c, ?> 2ab cos 2^ + cos 20, \{j; («^ - 6^) and y = a sin ^ + 6 + h^'+y-] = (.r^ + http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight //,2 - sin 2^ '2c-. sin 2^, i-'V-. 368/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELIMINATION. XXV.] 18. If tan {6 -a) shew that 19. If a cos2a + 6cos2^=c, i= a-b,, aud 7-; .f shew d^ = a (sin + c- - 3^ - sin tliat (.r^ cos 2a 2rtc and y = a (9), 325 +y^) (2«^ — = Ir. - cos (cos ^ .3^), - 2^')'' = 4a*.r'^. -^'^ Eliminate ^ from the equations: 20. xcoad„ 21. 22. a /-5 ^ cos x ^md-y .I'COS^ 23. r^\\\6=acosi6, 1/ —= wsiii^ +'-^-^ b If cos (a — 3^) +V , = m cos^ «i- Eliminate 6 and 24. tan 6 + tan 25. sin ^ 26. « sin 27. If + sin ^ + -cos (9 = <^ </) + A cos ^ sin ^ + cos =-, = a, -1- 30. «cos- ^ +6 X cos 5+// csin 5 H. K. 1:. = ??i sec- ^ 2 6 + — tan- 6-n sin 5 = 2a ^/3, {6 „ sin^ v.. t;-^ /)-icos-^. 0, = asiu(5 + 0), (^ = a. « tan , d~b tan ^-f/j ;iiid 0. = n, . ^-</) ^;, tan- = «, a. : = 0. x cos ((9 sin (5 sin (f) = a. asin- 5 + 6 cos ^^ = «, sin-0, tan- — (p)+y (f) = 1, = — {a + b^ cos- 0, m X cos 31. 4) = 6, + cot cot ^ d between the equations sin2^ =m 3^) sin </)= +f h = ,^ (^ + b sin-' b- ah {ah — Eliminate ^ and 29. (fj cost/) a- shew that - + cot <^ =y, (( + tan sin (a „ , + w cos a = 2. cot ^ .r, shew that If tan ^ and ^, /— . . from the equations = a, -''' ^ = rs^ — i'^. 1 9-, -I — — -sni^-ycos^=\'<r-sin2^ + cos- ^ = « cos- '' fr 28. ) . . ^, shew that ain- 6 6 ->- = ^ '-'JS 3 \/x- cos^ = 2asin .rsin (J+y = /j + 0) +y sin (<? + 0) = 4a, - 0) = 2a. sin 5, T. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight cos5 — cos0 = 2)Ji. 23 369/467 , 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 826 [c'HAP. Application of Trigonometry to the Theory of Equations. 326. In the Theory of Equations it is shewn that the solution of any cubic equation may be made to depend on the solution In certain cases of a cubic equation of the form a?-\-ax-\-h = 0. the solution is very conveniently obtained by Trigonometry. Consider the equation 327. — A-^ we —r= the identity cos 3^ = 4 cos Let A-=y cos - 3 cos 4 \ 4 and ?/=^ + » / — - , since y is a positive quantity; then from is positive ~ 4 whence cos Hence the values that _ r p (1), (3). (3) are identical, cos 3^ ) (21 = cos3(9--^cos(9--^, yi y1 If the equations (2) 1 ^, — =0 cos 3^ where y 6, ^ 3 cos^^--cos^ liave ( j and '/'represents a positive quantity. in wliicli each of the letters From g'.'P we have 7^ = • t, so that and ; _ ~ hllr'^ V —^ 3^=^/27r^ 64^ ^ ' . of 6 are real if 27r'^<4g'3 ; (0' <( )'• is, if Let a be the smallest angle whose cosine then cos 3^ = cos a whence 3^ = ; Thus the values a 2TO7r is equal to w/27r2 —3- ± n. of cos 6 are COS , cos —-+—a 27r , cos http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 2;r-a - - - n^, . . [See Art. or> 4.J 370/467 . 8/12/2019 . Elementary Trigonometry by Hall and Knight 327 AVPLICATION TO THE THEORY OF KQUATIONS. XXV.] ^=ycos^= ^-^cosd, But —qx-r~Q and therefore the roots oi afi Aa ^f 328. we may are 2rr-a Ao Ao27r + a ^_cos-— ^^COS-3-, cosa, Following the method explained in the preceding use the identity 3 ^ , . 3d sin ^ . 4 article, „ 4 to obtain the solution of the equation .^•^ — qx 4- r = 0, each of the quantities represented by q and r being positive. Example. + 8 = 0. Solve the equation x^ - 12.r sjn 30 3 We have —y sin 3 .•. T r, 12 — — = -^ = sin 30 4 and 8 y* ; Suppose that sin y- = —5 ; 1 -; 8 n the values = 50°, = 10°, 1 o^ = -sin 30 2 • whence + {- ) ; then 30°. 0, 1, 2, 3, = 130°, measure 4 we obtain = 170°, = 250°; n the values 5, 6, 7, ... it will easily be seen are equal to some one of the three quantities and by further ascribing that the values of sin positive; then whence « =4 , ; is + -^ ^ 0. d is estimated in sexagesimal ascribing to to sin 10°, .T where y 6, \r 30 = «. 180° But 4 8 12 sin^ d By —= 4 sm^'^-^rSin^H In the given equation put x ^ 0. -. =?/ sin 0«=4 sin 0, 4 sin 10°, sin 50°. - sin 70°. and therefore the roots are 4 sin 50°, -4 sin 70°. 23 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 2 371/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 328 [cHAP. Application of the Theory of Equations to Trigonometry. In the Theory of Equations 329. whose roots are a^, Og) <^3) > it is shewn that the equation ^^n ^^ {x-ai)(x-a2){s-a^) ^»-;S'i^-i or where + ^2-=«^ -2->S3^«-3 (.?7-a„)=0, + {-l)^Sn = 0, + *S'j = sum of the roots ^2 = sum of the products of the roots taken two at a time = sum of the products of the roots taken *S'3 three at a time; >S'„ = product of the roots. [See Hall and Knight's Higher Algebra, Art. 538 and Art. 539.] Example 1. If equation shew that (i) a, /3, y are the values 6 which satisfy atanS0 + (2a-.r)tan^ + ?/ = O (i) if (ii) if From + tan ^ = tau a tan /3 = tan a 7t, A-, .•. then But a tan*7 .-. From + 7/- + {2,a (2a.-;i;) (1), - x) ak ^ we have from + tan ^ + tan 7 — 7 = 0, 7n- tan the then ah^+{2a-x)}i = y; the theory of equations, tan a (ii) of or tan 7= = a^k . (1), ; -h. tan7 + 9/ = 0; ah^Jr(2a-x)h-y-0. we have from the theory of equations, tan a. y= B tan tau •^ (1), ?/ - rt w .-. A;tan7= -- , or tan7= - a Substitutinf? in a tan^ 7+ ?/ ak ^. {2a - x) tan 7 + ^ = 0, we have -^,-(2a-x)4 + y = 0; a^k .: ?/2 ^ ' ak •' + (2a-.r)rtA-2-«2/,»=0. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 372/467 8/12/2019 Elementary Trigonometry by Hall and Knight APPLICATION OF THE THEORY OF EQUATIONS. XXV.J Example Shew 2. cos- a that ,/2r + cos-' q \ „/27r \ + + ) 4 cos* ^ - 3 cos 9 \ = cos 3^ — k ) 3 ~ 9 ; = 0. 4 4 The '^ ; cos»6I-tCos<^--' .-. \ ^ ^'^^' cos Sd=^k Suppose that then 329 roots of this cubic in cos 6 are cos a, COS -TT ( + ^^^ ' ) ^'^^ ^ ^ ) I ' any angle which satisfies the equation cos3a=/i;. shortness, denote the roots by a,h,c; then where a is a + 6^ + c'^ ={a + .,/27r a s- 330. If 5d = + cos- ( -o- h + cf- \ + + I + ca + ab) 2 {he „/27r COE- ~ ( ^ 2tin, where n is any For \ '^ ) 3 = 2 we have integer, 3S=2n7r-26; The values of sin ^ found „ Stt . -sin sin 3(9= .-. this equation are fi'ona 47r . 2(9. Gtt . 8rr . , 0, sin sm -y- — , sin — sui , — , It will being obtained by giving to 7i the values 0, 1, 2, 3, 4. easily be seen that no new values of sin 6 are obtained by ascribing to n the values 5, 6, 7, ,, , But . sni .477 —- = - sin — = - sm 67r . - u O . and TT Stt . sm — o = - O , Stt sin o — ; hence rejecting the zero solution, the values of sin 6 found from the equation sin 36= — sin 20 are _L + sm • '^ - , J and ^^ + sm ~ j_ • http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight . 373/467 . 8/12/2019 , Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 330 If we pvit siu d = x, .r, and thus removing the solution - 4^2)2 a quadratic in .^'=0, we have = 4 (1_ ^-2)^ -20^H 5 = 0. 16^7* or is 26 becomes - 4x^ =—2x^/1 — x'^. (3 This 3^= - sin the equation sin 3.r Dividing by [CHAP. and as we have just seen the vahies x^, oi x^ are - and sm* 5 5 sin-' From „7r ^2iT . , 5 5 „7r 20 5 16 4' 5 „27r sm2-sm2- = . Shew Example. . = 2mr, ^-g. where n _ + smy + sm y = ~ ^7. 2ir is 47r . any Bin The values . that sm integer, 4^= -sinSe. . sm a;, — = - sm — 6jr (1 - (1 - Stt . 4^= - ±sm — sin 3^ becomes Ji-x^=Ax^ - 2x2) rejecting the solution 16 The ±sin— the equation sin 4x or , ±sm— suice = ,„ we have , If sin ^ 1 Stt . of sin 9 found from this equation are 0, whence wc have the tiieory of quadratic equations, . If 16 . a; = 0, we 3a; obtain - a;2) = 16x* - 24x2 + 64x«-112x* + 56x2 -7 = 4a;2 + 4x*) (1 9, (i). values of x^ found from this equation are sm2 — 2ir . , o Bm2 4ir y . , http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight „ sm2 Sir y ; 374/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight 27r iw iir . But (/ sm y+ 67r\ 27r COS—- COSy / + j — . y+ Btt —= ^ 47r Stt 112 = sm sin COSy- I . .Bit '2ir . „87r sin- -^ + sin sill - - 1 ,47r . y + sin- y sm- hence 331 THE THEORY OF EQUATIONS. APl'LICATION OF XXV.] . sm 7 y 127r\] iv / COS y - CCS ^ — IOttX COS-y- + 1 I = 0. — + sm y + sin Sir\y = sm- •lir Sin iir . .,27r _- . . 1 331. If 'id = -2i'Tr, 1 -— + sm -^ = .iir + Sin 2ir -— where zi cos .•. By . Stt . . sm giving to n the vahies any i.s j: . .^Stt 7 -^ ,^ V' integer, 4^= cos 0, 1, 2, 3, .,47r + sm- -= + sm- y - ' • we have 3^. the values of cos^ obtained from this equation are n the values cos Now and therefore Stt TfT = 4, 5, 6, y 7, ; IOtt 67r COS cos-^. COS^;;-, no new values of cos 6 are found by It will easily be seen that ascribing to Btt 47r 2jr cos—, 1, , cos -^r- = for 47r COS cos4^ = 8cos^ ^-8cos- ^-fl, if a: = cos 6, the equation cos 4^ = cos 30 becomes 8x^-4x^-8x^ + Zx' + l=0. or Removing the x-\, which corresponds factor to the root cos^=l, we obtain the roots of which equation are 47r 2ir cos y , cos y , cos http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight Qtt y . 375/467 ' 8/12/2019 . . Elementary Trigonometry by Hall and Knight ELEMENTARY TRlUONOJiETKY. 332 Example Find the values of 1. tau- - If 70 — mr, + + tan- -^ where n tan- -^ = t, writing tan and tan — tan — - tan — any integer, we have is tan By [cHAI'. 4^= -tan 30. becomes this equation - _ _ 3t-t^ ~ l-6t2+(4- 1 - 3(it 4f» t«-2lt* + 35t--7 or The roots of this cubic in ,., — tan- are t- ... 2t -—- tan- , 7 Stt „ --- tan- , 7 TT 7 '27r „ Stt = 21, tau- yr+ tan- -=-+ tan- —., .. = 0. ., and tan , — tan ~ = tan =- .^, ^/7, the positive value of the square root being taken, since each of the on the left is positive. factors Example 2. Shew that ,7r cos* - + cos* .TT . and + sec*- 27r y + .nw cos* 2ir sec* -^ + , ^ = ~ = 416. Stt sec* -=- Ill I?, , _ ; „ 7 Let y denote any one of the quantities „ TT cos- then 'Iij — l + x, ., = , cos- 27r ,, -=- , Stt cos- ~- ; where x denotes one of the quantities — Air 2-ir cos - , cos -=- , cos Gtt -=- . 7 7 7 From Art. Sol, the equation 47r 27r cos „ , 7 is Sx- whose roots are cos „ fiTT , cos „- 7 7 + ix- - ix - 1 — http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight ; 376/467 , 8/12/2019 Elementary Trigonometry by Hall and Knight APPLICATION OF THE THBIOKY OF EQUATIONS. XXV.] whence by substituting x = '2(/ - „2ir cos- -=7 „ TT — cos-' follows that it 1, , 7 333 „ Sir cos- , -zr 7 are the roots of the equation 8 -1)3 + 4 {2y - (27/ or _ 80^2 + 647/3 .-. + cos-- l)-i cos-* - 4 squaring the 24 .,27r = 0, ; 3 -. and subtracting twice the of these equations fii'st 1 ~ + cos- - = ^ = - oT By - 1) -1 = 0. 24(/ ^cos--cos--=gj = and - (22/ second equation, we have TT , + cos* ^ 7 By putting z — -, wc , cos* 27r -=- 7 + 37r cos* -— , 13 -—, — 7 11) see that „ 27r „ TT - sec- 7 sec- , ' —- „ Stt sec- -=7 , 7 are the roots of the equation ^3 .-. sec* ^+ sec* To ~ + sec* ^= rj 7 7 332. _ 242- + 80z- 64 = 0; find cos hti cos b6-\- cos ^ and sin h6, (24)2 I \ _ (2 \ x 80) I = 416. we may proceed = 2 cos 3^ cos 26 = (4 cos^ ^ - 3 cos 6) (4 cos^ - <9 cos5^=16cos^^ — 20cos^ ^+-5cos . . as follows 2) ^. sin5^ + sin^ = 2sin 3^cos2i9 = (3 sin - 4 sin^ ^ - 20 sin 5(9 = 16 (9 .• . sin-'^ ^) (2 - sin-'' 4 sin'-' 6-\-b sin 6) ; ^. It is easy to prove that cos 6^ = 32 cos tuid sin ^ 6^ = cos 6 (32 - 48 sin-'' ros* ^ B 32 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight + 18 cos- siir' 6 i 6* - 6 sin I 0). 377/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETKV. 334 EXAMPLES. XXV. Solve the following equations 1. ^-3j,--1=0. 3. .r'_3^._^/3 5. 8a^x^-eax + 7. If sin a and sin If tan a 8. shew that and if (I, tan 8.r^-6.i+V2 .r^* 6 sin ^ /3, = 0, l, if 2a=^ cos 3.1 - 0. + c = 0, + 2b'^ + 3ab + ac=0, a + c = ± 6. (2a - ft) if a tan a + b tan/3 = 26, + c (a - 6)2 tan y are the roots of the equation atan='^ + (2a-^)tan(9+3/ = 0, j3, 5a, x±^ = 3a. shew that cos y are the roots of the equation ^ + cos cos a (cos = 0. /3, cos^^ + acos^^ + ftcos and - 3a%- - = 0. then a~ then and a (tan2a+tan2/3) = 2^ — If cos a, cos 10. c b^ If tan 4. and tan /3 are the roots of the equation atan^^-fe tan B + 9. 4- a = a sin (2) if c sin a.-'-3.r 6. a + 2sin/3 = (1) if sin + l=U. 2. are the roots of the equation /3 a sin^ ^ shew that c. : = o. 2sm3A=(l [CHAI', = 2b, y) 6+c = 0, prove that abc + 25^ + c^ = 0. Pro\«e the following identities + sec — 11. sec a 12. sin2a+sin2('y cosec 13. ( -r a j + sec 14. cosec^ 15. cos 27r + a ( -^ - + cosec2 -- — + cos o 477 5- —-o 1 = - 3 sec 3a. + aj + sin2(^y + aj = -. cosec a ( = cosec ) + ( -— + =3 a + cosec 3u. j = 4. - 1 2 , ' ^ '^ *^ ^ http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 47r 27r y ^^'^ 5 1 ~4 378/467 , , 8/12/2019 Elementary Trigonometry by Hall and Knight XXV.] 16. APPLICATION OF THK THEORY OF EQUATIONS. Form the eqiiatiou whose roots are cos = (1)/ \ , -=- cos Form cos , sin2— 37r 14' .^SjT siii^ ^' . , 14' 14 the equation whose roots are „ TT . , . sin'= , Stt -- / and shew that 2 Form ,, cos ,-, =j^ and 2 lo ' Form — ; —= cosec* n=l Gtt 47r 2n- cosy, cos—, , TT cos-, cos (2) — — 5ir 37r , cos 32. < , TT and shew that Form cos cos , — Stt — ; Ttt . the equation W'hose roots are , 27r „ COS2-, COS-y, C0S2 — Stt „ , n=l 10 y ,1 47r C0S2— 2 cos*— =7^, and 2 =1 sec* -^ = 1120. 9 the equation whose roots are tan2 - and shew that 21. • 7 the equation whose roots are (1) 20. Stt — , . sm- , / sin* -^ n=l 19. , / sin2'— ^ , sm-' ^ 18. -;r / / 2) 17. 335 cot^ , tan- - + cot^ y — , —+ tan^ — cot^ -3- y y , tan- — , = 9. Prove that 1 ( \ ) ; cosec^ 7 + coaec^ —- + cosec* -p- = 8 27r (2) 7 7 37r 47r Stt TT cos-cos- cos^-^cos-cos- = http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 1 32. 379/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 336 MISCELLANEOUS EXAMPLES. If 1. tana + rt 4. cos* a a b — Shew true _ 1 a+b cos^ a 1 h jtan ^tan (j 2 \4 [ - f)| 2/J = tan- (î^«^ ) \cosa + sin/3/ . If the equation sin^ is —— that tan- 2 sin* a sin® a , prove that 3. + 6)tan I. acoafi — bcosa. prove that , tan^ = (a ft [CHAP. when n = a cos^'* prove that l, a be true when n will it is any positive integer. 5. a cos ^ + 6 If prove that 6. 4a^b^ ^ + 6 sin^ ^ = c, + {b — c){a — c){a — b)^ = 0. - y) - j8) : - y) = - 11 (i) 2 sin (/3 (ii) 2 sin asin (j8-y) cos(/3 + y-a) = 0; 2 sin a sin If P prove that (1) (2) A cos (a (^-y) PAB = sin it Two sin + y-a) = 2n sin^-y). ABC, such that = cot ^ -f cot 5 + cot C cosec'^ b) = cosec^ A + cosec^ B + cosec^ cot O - y) L PBC= L PCA = o), co hill of inclination railway on cos (a be a point within a triangle L 8. cos'-^ Prove the following identities (iii) 7. ^=c and a sin 1 in 169 faces West. which runs S.E. has an inclination of C. Shew that a 1 in 239. a are inclined to one At noon the breadtli of their shadows another at an angle a. shew that the altitude 6 of the sun is given by the are b and c 9. vertical walls of equal height : equation ii- sin- y cot''^ Q— })^-\- c- http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight + 2bo cos y. 380/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight MISCELLANEOUS EXAMPLES. {Including Chapter I. si I— K. lT/.'> Express in degrees aud luiuvites and also in grades the vertical angle of an isosceles triangle in which each of the angles at the base is twelve times the vertical angle. 1. 2. The angles of a triangle are as 4:5:6. Express them in radians. ^ 3. T, X, i Prove that — cot A — tan A — —— = 1-2^ cot 4. tan If A is , j- o , sm-.4. A + tan A an acute angle and .sin J = 5 v^; fi d the value of A + sec A 5. The adjacent sides of a parallelogram are and the included angle is 60% find the length diagonal to two places of decimals. A 1 5 ft. and 30 ft., the shorter (»f From high stands on the edge of a clift . a point in a horizontal plane through the foot of the clifl^', the angular elevations of the top and bottom of the tower are ob.served to be a and ^, where tann = l 26 and tanj3 = ri85. 6. tower 50 ft. Find the height of the 7. 60 ft. 8. cliff. Find the length of 10 degrees of a meridian upon a globe in diameter. The sine of an angle is to its cosine as 8 to 1 5, find their actual values. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 381/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELKMENTARY TRIGONOMETRY. 338 9. Find the values of 6 from the equation 4 sin2 ^ + ^/3= 2 1A 10. i Tr If tan a 11. Prove that 4 =— 15 „ ) , ,, (1 + ^3) sin (9. , „5sina find the value of 6 cos a 7cosa -— + — . (l+sin.4+cos^)2 = 2(l+sin.4)(l+cos 12. — . 3 sin a J). Simplify the expression 2 sec^^-l — sec*yl — 2 cosec^^ -f-cosec^ J, giving the result in terms of tan A. are sm a - cos a = -: prove that smn + COSa *^/2 sin ^ = sin a - cos a. o 13. T^ 14. Shew that 1 tlie 1 1 a 4. tan 6 . , the values of sin 45° - cos 45° + cos 60° sin 30° sec 45° - tan 45° , and cosec 45° + cot 45° same. Prove that the multiplier which will convert any number 15. centesimal seconds into English minutes is •0054. of 16. Prove the identities A — tan B cot 5- cot ^ tan (1) ^' 17. l+tan'-^g i+cot2e B tan cot -4 ' / 1-tan^ _ Y ~vi-cot^y Solve the equations : 3 (1) sin ^ + cosec <5=-y-; http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight (2) cos ^ + sec ^ = 2^. 382/467 8/12/2019 Elementary Trigonometry by Hall and Knight MISCKI;LANEOUS KXAMPLKS 339 K. A man running on a circular track at the rate of 10 miles an hour traverses an arc which subtends 56° at the centre 18. Find the diameter of the in 36 seconds. If 19. equilateral figure, shew AD tt = 22 -^^ . drawn perpendicular to BC, the base of an triangle, and BC=2m, find AD. Thence, from the is tliat cos2 60'' + cot2 30°=-, . 4 Pi'ove the identities 20. J +cos2 J) (tan2^ - = (tan2y| + 1) (sin'M - cosM). (1) (sin2 (2) sin2ocos2|3-cosâRin2^=sin2a sin2^(cosec2j3 — cosecâ). l) In a triangle, right-angled at C, find 21. a + c = 281, c and b, given that cosi5=-405. On a globe of 6 miles diameter an arc of 2 fur. 55 yds. measured find the radian measure of the angle subtended at 22. is Take circle. : the centre of the globe. If this was taken as the unit of measurement, how would a right angle be represented Shew 23. (1) that sin d cos tan ( 5 - (9 u - sin r (^- ^ , , ) 7 V2 ., cosec ^ + eos ( 5 = 7 soc ^| 1 - ^ v^ ) 1; , sec(^-. From a station two lighthouses A, B, are seen in direcN.E. respectively but if A were half as far off as tions N. and appear due W. from B. it really is, it would Compare the distances of A and B from the station. 24. ; Find the numerical value of 25. 3 tanMS and find - sin2 60° -\ cot2 30° +3 sec2 45° ; x from the equation cosec (90° -A)- x cos A cot (90° -A)= http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight sin (90° -A). 383/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 340 Prove the identities 26. (1) {siD.A-cosecAy + {cosiA — secAy — cotÂ+tanÂ-l; (2) (cot ^- 3) A (3 cot (9- 1) =4-5, find the vakie of ^ 27. If cot 28. Find two values of c-ot which ^+ 1 0). ^ ~ ''''^ ^ A +3 cos A ^' ^ 2 sin satisfy = cot + 2 (9 20° 17' at an arc subtends If 10 cot , 2 cos 29. = (3 co,sec2(9- cos the 0. centre of a circle whose radius is 6 inches, find in sexagesimal measure the angle it will subtend in a circle whose radius is 8 inches. Looking due South from the top of a cliff' the angles of depression of a rock and a life-buoy are found to be 4.5° and 60°. If these objects are known to be 110 vards apart, find the height 30. of the cliff.' Prove that 31. + cos J. 7 1 - cos A sec 1 -; 1 ^4 8 ( 1 ) sin'' What (9 - - 1 , + sec A Solve the equations 32. 2 cos ^ —4 , ., , cot^ A , = 4 , 1 r + sec A . : =5 5 tan- (2) ; .v - sec^ .,• =11. the difference in latitude of two places on the meridian whose distance apart is 11 inches on a globe 33. same T whose radius is is 5 feet Take 7r= ? 22 -;:r • I 34. Given that sec 25 A=-- -^ , find all the otlier Ti-igonometrical ratios of A. 35. Which of which impossible (1) (3) tli(^ following statements are possible, and ? 4sin2^ = 5 (2) ; (m^ -f n^) cosec = m^ — n'^ ; http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight (a^ (4) + h^) cm0 = 2ab sec ; = 2*375. 384/467 8/12/2019 Elementary Trigonometry by Hall and Knight MISCELLANEOUS EXAMPLES 341 K. Walking down a hill inclined to the horizon at an angle B a man observes an object in the horizontal plane whose angle Half way down the hill the angle of depresof depression is a. 36. sion is Prove that cot ^ = 2 cot a - cot ^. /3. {After Chapter XII.) II. In a triangle a=25>j2, perpendicular from C on c. 37. c = 50, (7=90° find B, h : and the Prove that 38. (2 sec A +3 sin ^4) (3 = (2 cosec cosec A -2 cos A) J +3 cos^) (3 39. Find the values of sin 960°, cosec 40. Find the equation all — 510°), tan 570°. the angles between 0° and 500° which satisfy tan''^^ The angle 41. ( sec .-1-2 sin A). = l. of elevation of the top of a steeple is 58 from a point in the same level as its base, and is 44° from a point 42 feet directly above the former point. Given that tan 58° = 1-6 and tan 44° = -965, shew that the height of the steeple is 105 ft. From 42. approximately. the formula i?a\A=1 tan 75°, 43. ( 1 44. and solve the equation sec- ^ find and tan 15 = 4 tan 6. Shew that + sec ^ + tan d){i-{- cosec 6 + cot S, = 2 (1 + tan ^ 4- cot ^ + sec ^ right-angled at C shew ABC sin^ A cos^ A a^ — b* In a triangle sinî/ cos'^ B Find all the angles less satisfy the equation 45. 2cos'-^-i II. ^, + cos 2^ K. i;. + cosec 6). that a^b^ tliaii four right angles which + sin^. T. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 24 385/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 342 46. JJeterniine the value of sin (270° 47. GiAên siiia and de(Uice .- -p, sin 5 tlie vabie of 4 . sin/3=-, find > ) when + n +/3) = ^r^^- —2 A——cossin^ — cos , —. ^4 .3^4 --, . .- sin If cos A — 'G. cos(«-t-/3), . , • . to a single ° 1 A= — /3 ^ , 1 term and trace the changes of the expression in sign and magnitude from 0° to 180°. 49. sin 79 \/2 (4.5° Kecluce 48. =— +J as A increases drawing a diagram to explain find tan J. the two values. From 50. balloon a over vertically a straight road, the angles of depression of two consecutive milestones are observed to be 45° and 60° find the height of the balloon. ; Find the value of 51. (1 cot2 ) - 2 cos2 ^ - ^ 2 sec- 180° sin 0° (2) 52. g sec- - cos - ^ 27r 4 sin^ ^ + cosec — : . Prove the following identities .sin*a (1) l + + 2sin'- ail \ ., ) cosec- aj = l-co.s'a: Q- tan2 (2) ^= cosec 2^; y l-tan-'(^^-^j (;i) 53. cos 10° + sin 40° = ^/3 sin If bi-M\ d = a, find the value of a sin a sin 54. 70°. 6-b cos ^ + 6 cos B S' Prove; that 4 cos J 8 - 3 sec lb' =2 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight tan 18 . 386/467 8/12/2019 Elementary Trigonometry by Hall and Knight MISCELLA-NEOUS EXAJU'LKS 55. 343 K. Find the values of tan (-240°), cos 3360', cot (-840°). Prove also that • sin — - cos - + cos Stt it TT — = sec 2n . A railway train is travelling on a curve of half-a-mile radius at the rate of 20 miles an horn- through what angle has 56. : it turned in 10 seconds? 57. If sec a= Take =— tt . 13 , , find the value of 2-3cota 4 58. ^/sec2 a- 1 Prove 2-2tan Jcot24 = sec2,-l; (1) --^ - cos (2) 59. —9 }/ ( -+ cos i^_^+,.2=a When A+B->rC=\ 80°, cos ,. A ,„, (2) sin A 5 sin simplify +B) cos (C+B C- sin (^ + 5) cos 5 C sinCsin^ cos (C+ 5)' cos (7+ cos {A cos ^ sin cos .4 ^^ ( cos ) (7 1 sin^sin_5' 100 feet high stands vertically at the centre of a horizontal equilateral triangle if each side of the triangle subtends an angle of 60° at the top of the flagstaff, find the side of the triangle. 60. flagstaff : 61. Prove that the product of + sin ^) + cos ^ I +cos 6) sin 6(1- .sin + cos ^ (1 - cos 6) sin ^ ;i'i(l is equal to sin ( I ( (9) 26. 24—2 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 387/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 344 Shew that 62. (l-sin^)(l-«in(/>) = |sin^-cos^r. Prove that the value of 63. + ^) - sin (a sin (a - 6) cos03-^)-cos(/3+^) is the same for li 64. all values of 6. A+B-\-C=^\ 80°, prove that A cos COS-^ B-C — 1- cos B — COS C-A — = sin ^ + sin B + sin a If tan - 65. = cosec 6 — sin cos 1- C COS — A-B C. shew that 6., Q cos2^ = cos36^ or cos 108 . A man 66. stands at a point and X on the bank XY of a river and observes that the line joining to a point Z on the opposite bank makes with JTlân angle of 30^ He then goes 200 yards along the bank to Y and finds that YZ makes with YX an angle of 60°. Find the breadth with straight X parallel sides, of the river. found that the driving wheel of a bicycle, 32 inches in diameter, makes very nearly 1000 revolutions in travelling Use this observation to calculate (to three places 2792w yards. of decimals) the ratio of the circumference of a circle to its 67. It is diametea\ 68. If a+/ii+y=|, prove that sin''^ 69. a + sin^ /3 + sin* + 2 -y sin a sin /3 sin y= 1. Prove that (1) (tan (2) sin- .1 A + tan cos'* 2^4) (cos A= ,-. lb + .1+ cos 3.1) = 2 .,^ o'l cos 2.1 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight , ,, 10 sin 3.4; cos 4vl - -^ cos G J. <oii 388/467 8/12/2019 Elementary Trigonometry by Hall and Knight MISCELLANEOUS EXAMPLES 70. U a = j^, 19' 71. If A sill find the value of +^ = 225°, 23a A cot B 2' Prove that cot 6 - tan 6 = 2 cot 26 tan 6 + 2 tan Simplify 74. Eliminate ^ .'r -^ =3 2(9 l+cot5 A sin : and hence shew that + 4 tan 4^ = cot ^ - l-,-^^- 73. sin 3a sinl6a+sin4a l+cot5 1+cot^ . S^J prove that cot 72. — K. /f^ 1+tan^ 8 cot 8^. . between the equations 4 -sin 3^, ?/ = cos 3^1+3 cos ^. A, B are stand on a horizontal plane. two points on the line joining the bases of the flagstaffs and between them. The angles of elevation of the tops of the flagstaffs as seen from A are 30° and 60°, and as seen from B, 60 If the length of AB is 30ft., find the heights of the and 45°. Two 75. flagstaffs and the distance between them. Prove the identities 76. ( A + cos2 i5 - 2 cos 4 cos 5 cos (^ + ^) = sin2 {A + B) 2 sin 5.4 - sin 3.4 - 3 sin .4 = 4 sin A cos^ .4(1-8 sin^ A cos2 1 A 77. is ; ). (2) which flagstaffs 3 inscribed in a circle the circumference of Find the number of inches in the length of a square feet. side, correct to is two places of decimals. i=-3183, v'2 Given = 1-4142. IT D are taken on the circumference of a circle so that the arcs AB, BC, and CD subtend respectively at Shew that the centre angles of 108°, 60°, and 36°. 78. Points A, B, C, AB=BC+ CD. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 389/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 346 79. Prove that cot 15° 80. From + cot 75° + cot 135° - cosec 30* = 1. the equations cot5(l+sin^) = 4m, cot^(l-sin^) = 4?i, derive the relation {m^ — n-)-=mn. Prove the identities 81. + a)cos-y = .sin(/3--y)co.s(a + ,S + y); (sec A + sec 3.-t = 2 sin A sec A sec Â (1) siu(a+/3)cos/3-sin(y (2) (tan 2^ - tan A) ) 82. Prove that cos 83. From 6° cos 66° cos 42° cos 78° the f(jruiula cot A= cot22°30' 84. An 1 cos — sin 2 A 2x1 -4- -. ^r-r- , =— . prove that = v/2 + l. observer on board a ship sailing due North at the' rate of ten miles an hour, sees a lighthouse in the East, and an lighthouse bears S.S.E. find in hour later notices that the same miles, to two places of decimals, the distance of the ship from ; the lighthouse at the first III. observation. (After Chapter 85. Prove that (1) sin.l sin(5-(7) (2) ^'''^^ d+ sin 2^ = i^cose + GOs2e- XVl) + sin7isin(6'-.-l) + sinC.sin(/l-^) = 0; sin 86. If n+^ + y = 0, cos a 87. I prove that + cos p + cos y = 4 cos g cos — COS 9 — I- n any triangle prove that con A+ , cosB-] http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight COS 6'= 0, 390/467 . 8/12/2019 . Elementary Trigonometry by Hall and Knight MISCELLANEOUS EXAMPLES xi — sec 2^ Prove that loga h Given logio3 = -47712, Tf .4 +^+ cot A <: 4- cosec 16 log,, = a L logjo 5400, B 1 = -90309, logio 8 = 90°, cot + logj c log,,, 2-4, 90. sin & ah X rove that 89. _ COS 6 If 88. 347 K. find tho values of tan 'Mf prove that cot (7= cot A cot 5 cot + C J, i?, C are in Arithmetical Progression, shew that this equation gives tlie value of cot 15°. and if 91. Shew that (1 4- sin 2.4 4- In a triangle where of the roots of the equation a, 92. „ ,, , ?), cos^.4 (1 A 4- sin 2.4). shew that are given, c is one — 2hv cos A+b'- a- = 0. x^ „„ =4 cos 2^)2 sin 9° = 93. Prove that 94. U A+B+C= 180°, . .-,„ sin 48 sin 12° -,„ . smSl . prove that co.s^4-cos^ + cos- = 4 cos 95. find ( 45° — ^j ] cos ( 45 -— \ C08 f 45° - j ) • = 9-6680265, X sin 27° 46' = 9-6682665, L sin Given 27° 45' Lsind =9-6682007, 6. 96. Prove that of their cosines twelfth of the is sum A, B, C are three angles such that the sum zero, the product of their cosines is oneof the cosines of SA, 3B, 3C. if http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 391/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 348 If 97. A the values of 98. be between 270° and 2A and tan — ,sin and 3fi0°, .sin J= - —7 , find . Solve the equation 2cot| = (l+cot(9y-, Hence find the value of tan 15°. Given logio2=:-.3010300, logi^ 360 = 2 5563025, find the logarithms of -04, 24, -6, and shew that log^ 30 = 4-90689. 99. 100. Prove that cos (a;—i/ vanishes — z) + cos when 101. .27+?/ If cot a {y — z — x) + cos +3 is (z — .r — y) — 4 cos x an odd multiple of a right = (^3 +,^2+ ^)i^ cos y cos z angle. ^ = (.^+^-1 + 1)5, cot (.t?-3+.r-24.^-i)i^ tany = shew that a + /3 = y. 102. Shew how to solve a right-angled triangle of which one acute angle and the opposite side are given. Apply this to the triangle in which the side is 28 and the 31*53' 26-8 , given log 2-8= -4471580, log 4-5 = -6532127, angle L cot 31 103. If tan prove that ° 53' A= = 10-2061805, ^ ^ 4-V3 and tan tan {A for 1' difl:: B^ ^^^ 4 + ^/3' — B) = -375. 104. The 105. Prove that cos A - sin sides of a triangle are x, y, its greatest angle. and = 2816. ^4 is a and \/x^+xi/ + i/'\ ftictctf- of cos 3^4 find + sin 3^1 . ; tliat cr>s2 A + cos'^ (a + —) + cos^ http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight (a ~ ~\ 392/467 . . 8/12/2019 Elementary Trigonometry by Hall and Knight MISOELLAKEOUS EXAMPLES 106. tan any Ill triangle, tan^=-, and if 34f) K. tan — = —;, find C. Shew also that, in such a triangle, 107. Simplify -j 108. cot ^ If + cot a= ( ^ b 40, - ^)l = ni, r jtan = 43, (^ „ w sin 1 tan (.1 A —% cos A A - i5) = ( 1 - 71) tan ^ Given log 5 = -69897, find log 200, log and also Z sin 30° and L cos 45°. (1 (2) 7, r-2-j-, sin^ 110. 111. • = tan5 = - If + ^)| ^ ( 6-03 -78031 -107210, log = = 9-6634465. tan 24° 44' 16 Z prove that - ^) + t-an find the value of A, given log 1-2S 109. a + c = 20. log\^62^, -02.'i, * Prove the identities: (sec 2 A - 2) cot (A - 30°) = (sec 2A + 2) 1+ cos 2a cos 2/3= 2 (cosâ cos-/3 tan (.4 + 30°) + sin-n siu'^jS). BC is In a triangle, 5=60°, t^=30°, 5(7=132 yards. 15°; find TZ) and the i)erjiroduced to Z> and the angle A = that on given BC, ,^'3 pendicular from A 1^^ approximately. 112. 113. DB= In any triangle prove that / (a 114. If r G .. + 6 + c) tan ^ = « the sides of A ,A cot ^ 2 a B , + a cot ^ - c cot triangle are 68 — ft., 75 ft., 77 ft. respectively, find the lea-st angle of the triangle, given log 2 = -30103, 7. cos 26° 34' = 9 -95 15389, http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight diff . for l' = 6.32. 393/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRTf40NOMF,TRY. 350 If sin .4 = -6 values of sin {A - 90°), 115. between 90° and lies - cosec (270° loga Given logjo 5 <^ = log„ b = -69897, X logb C X loge d. find logics, logg 10, log]o( -032)5. Prove that + A) + cos (60° ~A) = cos cos 105° + cos 1.5°. Deduce the value of •V l^'ind .-1 . the values of tan - from the equation cos 119. the A). cos (420° 118. 180°, find Prove that 116. 117. A and If sin ^4 : .r — sin a cot j3 sin *'= cos sin (2 A i / i cot (A +B) — n + B) = TIN ni : a. m, prove that —n J cot , .4. m+n A AB AD stands on a horizontal plane, and AC, is 17 ft. longer than If are the shadows at noon and 6 p.m. AC, and BC is 53ft., find the height of the tower and the 120. tower AD altitude of the sun at noon, iven tan 31° 48'= -62. 121. altitude at 6 p.m. is 45°; Prove that + sin (1) sin 8<9 (2) sinl8° 122. when the Given 2^ = 4sin '- cos '.^ cos 3^; + cosl8°=:V2cos27°. lo^- 36 -= 1 -.556302, 10^48 = 1-681241, find loir 40 /Y 15- '^' ^^ -V Given i = 9-5, ^=-5, .,4 = 144°, find the remaining angles; log3 = -477121.3, Xcot72° = 9-511776(), 123. giv(ui /. tan 1 6° 1 9' = 9-4664765, /. tan http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 1 6° 1 8' = 9-466(X)78. 394/467 8/12/2019 Elementary Trigonometry by Hall and Knight MISCELLANEOUS EXAMPLES In any triangle prove that 124. bcsmÂ=a^{cosA+cosBcosC); bccosA+cacosB+2abcosC=a^ + h^. (1) (2) If tan 125. f = 4 tan - , prove that B—a — tan 3 sin a 5 — 3 cos a Shew that 126. sill(36° + .^)-sin(36°- If 127. figure 351 K. why sin^ J)+,sin(7-2°-.l)-sin(72'' 2 = — -, find tan ^, + J) = sin J. and explain by means of a there are two \'alues. Prove that 128. (1) sm2A+cos2B = 2smC^+ A - b\ cos(^~ A~Bj (2) (sin d sin (f)) (cos (/> In any triangle, 129. (sin - A +smB + sin C) + cos ^) = 2 sin (^ - <^) cos^ —^ ; . if A+sinB- sin (sin C) =-- 3 sin A sin B, prove that C=60°. />' 130. Prove that log„ b x 131. If log 2001 132. If a = 7, b If tan c^= log,, = 3-3012471, — S, c to the middle point oi 133. log,, = 9, AC log 2 dx log,, b. = -3010.3, find shew that the length log 20-0075. of line joining is 7. A +sec .4=2, prove that sin A = 3^ when A is less than 90° 134. Prove that 3-4oos2/l + cos4.I 3 + 4CO.S 2 A + cos 4.1 = taii-' http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight .1. 395/467 . 8/12/2019 . Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 352 Shew that 135. sin3/l + cos34 sin 3^ u OJI — cos uuw 3^ oî X cos J. y cos 45°). ^ 5 = (.r +y) tan x tan J + ?/ tan prove that - tan (A i. If 136. —-— Given 137. log 3-5 find log 5, = -544068, and log log 7, = -511883, log 3-25 log 2-45 = -389166, 13. In a triangle, a = 384, 6 = 330, (7=90°; 138. angles 1+2 sin 2^ — ^ sin 1-2 Zrî am 2A find the other given ; = 1-0413927, log 20 = 1-3010300, L tan 49° 19' = 10-0656886 L tan 49° 20' = 10-0659441. log 11 If cos 6 139. = cos a cos , tan prove that /3, e-a e + a. —— tan — = tan-'„/3 ^ . Prove that 140. sin 6 sin sin 6 d) sin <i . ; cos^ + sin^ If 141. in 1 cos a sin 0- z- — triangle ^ = _ ^- sin cos + 6) c (a cos B — .r j^ eos0+sin^ —•/;((/ + <?) cos C — , _ prove that b = c. 142. Prove the identities ,,, cot4+cosec.4 (1) ^' -: tan^+8ec.4 (2) sin3 143. log 3 = . -A ,(nA\ J+sin3(l20° + J) (Calculate the value of -4771213, log 485.59 -jr + v .A 2' + 2/ cot— -7 =cot \4 I 1 sin3(240° + ^)= --sin3>4. 18 x -0015, having given = 4-6862697, http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight log 48560 = 4-6862787. 396/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight MISCELLANEOUS EXAMPLES 353 K. Find the other angles of a triangle when one angle is 144. 112° 4', the side opposite to it 573 yards long, and another side 394 yards long given ; log 573 = 2-7581546, Xcos22°4' = 9-9669614, log 394 == 2'5954962, Z sin 39° 35' = 98042757, Z sin 39° 36' = 9 -8044284. XVIII.) {After Chapter IV. In any triangle prove 145. cos A cos B cos bcosA + acosB acosC+ccosA ccosB+bcosC C _ ~ Given 146. log 119, logy and , If A, B, 147. = -8450980, and log 7 C log b'' + c^ 2abc ' 1-2304489, are the angles of a triangle, and find if B + smC) = sin A, tan^ - = tan )rove that = + 3^3 cos 6 (sin ) log 17 a^ — tan — Prove that the diameter of a circle is a mejin i)roportional between the lengths of the sides of the equilateral triangle and the regular hexagon that circumscribe it. 148. Given that the sides a and b of a triangle arc respectively 50v^5 feet and 150 ft., and that the angle opposite the side a is Also having 45°, find (without logarithms) the two values of c. 149. given log 3 find the 150. = -4771213, t.-os 2.r = 9-9770832, Zsin 71° 34' = 9-9771253, 71° 33' two values of the angle B. Prove that 2 cos Thence X sin find the cosec 3.r 2a; sum cosec 3x = cosec w — cosec n terms to + cos (2 . Z.v) 3j;. of the series cosec X^.v + cos http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight (2 . 3-.(-) cosec 3'U' + . . .. 397/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TllIGONOMETKT. 354 Prove the identities 151. : (1) cosÂ + fimÂcos2B = cos^B + sm^Bcoii2A; (2) sin 33° + cos 63° = cos 3°. the positive angles less than two which satisfy the equation Find 152. riglit all tan*.tt-4tan2.1 angles + 3 = 0. Prove that 153. cot — ;r The tangents 154. of 4 sin e 3$ 3 cot - l+2cos^' 3 two angles of a triangle are - Find the tangent of the third respectively. and , —5 and the angle, Also find the third angle cosine of each angle of the triangle. to the nearest second, having given = 1-5185139, log 56= 1-7481880, 59° 29' = 10-2295627, DifF. for l' = 2888. log 33 L tan a triangle If in 155. («2 shew that the If 156. sin triangle and R {A-B) = (a^ - 6^) is sin {A + B), either isosceles or right-angled. are the radii of the in-circle and circuin- a triangle, prove that circle of cos^ 8/Ji *• + 62) — + cos- — -f cos^ g [ = '2bc + 2ca +2ah — a'^- b'^ — c-. In any triangle prove that 157. cot B„ + cos 5 ^ + ——7^ = cot 6 J sin C cos A sm B cos A . C —cos ^ , , -. Given log 6 = -7781 51, log 4-4 = -643453, log 158. 1-8 = -255273, find log 2, log 3, log IT. Prove the identities 159. ( ; 1 {2) sin 3.1 =sin (sin 2.1 - sin .1 (2 cos 2 J - I ) tan (60° 2B) tan (J +/*')- 2 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight + .1 (shi- .1 - ) - tan (60° -A); sin- B). 398/467 8/12/2019 Elementary Trigonometry by Hall and Knight MISCELLANEOUS EXAMPLES Find the greatest angle of a triangle whose sides and 214 feet respectively given 160. 183, 105, log 82 = 1 -9138139, log 101 log 113 = 2-0043214, = 2-0530784, A log 296 = 2-4712917, = 9-8360513, 34°27' = 9-8363221. 34° 26' Ztau Ztan and a regular octagon have the same perimeter circle com} lare their areas, given ^'2 = 1-414, 7r ; = 3-1416. a triangle he in arithmetical progression, a be the least side, then 4o-3?> 162. if ai'e ; 161. and 355 K. If the sides of cos 163. If a sin {6 —- , A= + a)=b sin . — 2c (^ + /3), prove that acosa— 6cos/3 r-b sm /3 — a sni o ^ cot B = ,—T— ^ 164. . In the ambiguous case shew that the circum-circles of the two triangles are equal. 165. From a point A on a level plain the angle of elevation and its direction South and from a place B, which is c yards South of A ou the plain, the kite is seen due North at an angle of elevation /3. Find the distance of the kite from ^1 and its height above the gi-ound. of a kite 166. is a, ; If a + /3 + y = 27r, express cos o + cos /3 + cosy+ 1 in tlio form of a product. 167. cos Prove that 10..4 + cos 8^1 + 3 cos Â + 3 cos 2 A = 8 cos A 168. In any triangle shew that 169. If tan- ^ 170. = 2 tan- + J, cos2^ + (^ cos^ 3^1. then sin'-'fj ) = 0. I'ruve that tan .1 tan(60 + J)tan(120' + Jj=- -tan3J. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 399/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 356 A=2B, 171. If in a triangle 172. Shew that the length inscribed in a circle circle as J3 inscribed in the to that of a square •J'i- = 76, and ^ = 6° 37' 24 L tan 3° 18' 42 y—+B\ = -— , tan same ~ , find the other angles; given = 8-7624069, Ztan =-30103, log 2 b). of a side of an equilateral triangle In any triangle prove that tan 173. If 3c : is then a^=^b{c + = 9-1603083, 8° 13' 50 diff. for 10 1486. = 174. If ABC, and if D A be the middle point of the side BG of a triangle be the area of the triangle, prove that cot and A DB = 7 . 4A 175. Prove that tan 20° tan 40 tan 80^ = tan 176. If, in a triangle, 6 = ^3 + 1, c; = 2, and 60°. .1 = 30°, find B, C, a. 177. Pi'ove that the rectangle contained by the diameters of the circumscribed and inscribed circles of a triangle is equal to 2abc a + b + c' 178. Solve the triangle 7vsin 81° 47' log 2 =30103, log 3 = -477121 3, log 7 = -8450980, .__ ,,. 179. II when a = 7, diff. 1' = Sj'S, ^1 = 30 ; given = 9-9955188, = 183. — sia2a+sin2a' ^^ r— , sni2^ = ^ -,, . t.Mi ( [ + /:<) - . + Sin 2a sm , 1 prove that for b ±tan ('[ + http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight ) 2a tiin (4 + «')• 400/467 8/12/2019 Elementary Trigonometry by Hall and Knight MISCELLANEOUS EXAMPLES 357 K. On a plain at some distance from its base, a mountain At a station lying 3 miles is found to have an elevation of 28°. 77 yards further away from the mountain the angle is reduced 180. 16°. Find the height of the mountain in feet. log 1-6071= -2060, Z sin 16° = 9-4403, to X sin A —+ B ,,, (1) ^ (2) 4 cos* tan If 182. ^ tan 2 A-B 2smB 2 cos^+cos5' - sin 2A A+B+C = '^, and cos A + cos C= 2 J. sin* -4 A + tan — tan or else .14-^^ is C „ — C=4sin — With the usual notation he ca ah rj r, ?-3 The If « (2) C V ^'^^ * in = 4090, 9 sin (45° triangle, c a a b [a a h h c c = 3850, c = -7690448, log 1-7855 = = ) c log 5-8755 15' any ^ ^ „ \h circle in A\ b - 45° ( bisector of the angle X cos 32° 1 cos B, tt. sin a- ( ^^*^ ( sin 4^. that in any triangle Shew and the circumscribed 187. A / ^ an odd multiple of cos>4-fcosi?-sin 186. = 9-3179. =4: cos 2^1 1 185. 12° A shew that 184. Z sin Prove that 181. 183. = 9-6716, 28° A -^ ). prove that ) J meets the side BC in D shew that sec— = 38 11 , find A , gi ven log 3-85 = -5854607, log 3-811 :2517599, 9-9272306, Z cos 32° 16' -5810389, = 9-9271509. = Prove that cosec ^ - cot ^ = cos (15° -a) 1 + 3 cosec2 ^ cot- ^ sec 15° -sin (15°-a)cosec H. K. K. T. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 15°= 4 sin a. 25 401/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 402/467 8/12/2019 Elementary Trigonometry by Hall and Knight MISCELLANKOUS EXAMl'1-K.S 359 K. the perpendiculars from the angular points on the sides of a triangle, prove that 196. If i'l, p-ii s-re Ps SB?='^-; (1) PiP-.P^ A-, Ih- (2) = i, + 2.--l-cosC^ P1P2 P2 Pi' Find the perimeter of a regular cpiiudecagon circumscribed about a circle whose area is 1386 sq. ft. given 197. ; tan 12° = -213. The top of a pole, placed against a wall at an angle a with the horizon, just touches the coping, and when its foot is 198. moved a is )3, it from the wall, and feet further rests on the sill 1 ^s (1) 201. (1) {2) 202. = a cot sill , n any triangle prove that ah — r^^g 200. angle of inclination window: prove that the perpen- of a dicular distance from the coping to the 199. its _bc — r^r^ ~ ^1 ca ~ — /-g/'j r^ ' Prove that cos-ilJ = 2sin-i?; (2) 3 tan-il = tan-ig. Prove the identities (I tan .l cos 2 A + sec J)cot — = (cot.-l+cosec J)tan( 45° + ^) + cos 25-4 sin (45'' - A) sin (45 - B) + B) = Hm2{A + B). cos (.-1 Given log 3 = -4771213, log 7 = -8450980, Z sin 25f = 9-6373733; shew that the perimeter of a regular figure of seven sides is greater than 3 times the diameter of the circle circumscribing the figure. 203. If tan = -——y cot — , in any triangle, prove that C sm. c={a + b) ^ ^ ^. coscfj 25—2 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 403/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 404/467 , 8/12/2019 Elementary Trigonometry by Hall and Knight MISCELLANEOUS EXAMPLES 1+m cos(^ — a) tan ( T ^^ = I prove that ' ' cot (H- Solve the equations 212. (1) sin 56 (2) (1 If 213. a+h= = 1^'\ ^^^^^± ) If 211. - sin 3^= ^2 - tan + sin d) (1 COS.4+COS cos 46 26) ; = 1 + tan C /? 361 K. = 4 sin^- 6. any in triangle, prove that 2c. A 214. flagstaff standing on the top of a tower 80 feet high subtends an angle tau~i - at a point 100 feet from the foot of the tower find the height of the flagstaff. : 215. Prove that + cot-i8 + cot-il8 = cot-i3; (1) cot-i7 (2) 4tan-J-tan--L==-. 216. between 217. If 2 sin (8?i-f 3) g ^ -= - Vl +sin A + \l\- sin A, shew that A lies and (8h + 5)-. Prove that «inM^ + ,^)-sinM'^-n = :^sin^. ,8 218. lftan(9 = , 1 prove t)iat 2/ -x ^~— cos 1 ^ d) sin 6 ^2 V« 2/ itxn (b = ~ ~ ^ cos 1 \-v .. $ ' _x ,sin'^>~ y http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 405/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY, 362 Solve the equation 219. tan-i ta (.r + l) + tan~i (^- = tan~i — l) oi ; and prove that sec^ (tan + cosec2(cot'î3) = lo. Prove that in any triangle 220. 10^ sin 1 .u . .u also that the is i2) + sin 10^ + sin 10C= 4 sum sin 5 A sin f .u , , or the cotangents 5B sin 5C bn + C —^— —^+—B — — ^ f57r + ^ oi 57r , , equal to their product. If dy, d^, Jj ai'B the diameters of 221. circles, the three escribed shew that dyd,^ + d<^d^ + do^d]^ = (a + + cf. ft AB To determine the breadth 222. C oi a ravine an observer AB produced through i?, He then and then walks 100 yards at right angles to this line. places himself at in the straight line finds that subtend angles of 15° and 25° at his AB andofBG the ravine, given Find the breadth L cos 25° = 9-9572757, X cos 75° = 9-41 29962, log 37279 = 4-5714643, cos 40° log 3728 = 9-8842540, = 3-5714759. Prove that 223. (1 L eye. -cos 224. 6) {sec (9 + cosec (1 +sec (9)}2 = 2 sec2(9(l +sin 6). If in a triangle 6^=60°, prove that 1 a-\-G 225. e + 3 1 6 +c a + 6+c' Prove that 2 cos =^= -„ 2 V2+n/2 + V N/2-f-2cos^; the symbol indicating the extraction of the square root being i'e2)eated n times. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 406/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight MISCELIiANEOUS EXAMPLES m tan {a — 6)_ If 226. 6 — ^\a 2 then [ , angle = - 6) ' ( sides of a triangle are such that 1 that prove ' (a tau~i \n-\-m tan a ah The 227. — n tan 6 ~ cos2 cos^^ 363 K. ^2 + + »l%2 A =2 tan ~ ^ c (1 ,i,2 — 5 = 2 tan ~ n , _ ^2) ^ ( 1 + ^2) ' mw, and the area of the tri- mnhc m^ + n^ A h feet high placed on the top of a tower I feet high subtends the same angle ^ at two points a feet apart If 6 be the in a horizontal line through the foot of the tower. angle subtended by the line a at the top of the flagstaff, shew that 228. flagstaff A = a sin ^ cosec 229. 230. each side 2^ = acosec^ (cos ^-sin/3). Prove that A is - tan - 2 2 -I- - tan - — th '' - cot - - cot 4 4 ; shew that ~ side is eq ^ ual to sec At what distance will 6. inscribed in a circle such that is of the radius by each = 4 4 regular polygon centre subtended 231. and ^, the angle at the —3— ^ •- 2m- - 1 an inch subtend an angle of one second ? 232. If tan iy = 4tan~ia', find y as an algebraical function of x. Hence prove that tan 22° 30' is .r*- 6.^2+ a root of the equation 1=0. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 407/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 364 233. If cos 20 = 240 tana find ^^777; , 2o9 and explain double the If 6, (}) be the greatest and least angles of a triangle, 234. sides of which are in Arithmetical Progression, shew that the 4(1 235. cos ^) (1 sin (/)) = cos d + cos <p. : l — ^3. In any triangle prove that (1) sin 3.4 sin (2) a^ sin : : a, /3 + sin 35 b^ sin (C-^) +sin3Csin(^-5) = 0; {C-A)+c3 sin (A -B)^0. sin F is n. a triangle and a point If the angle CPB is (in If (5 -C) {B - C) + ABC is AP BP=m 238. 4^- cos sin ^; = tan X - v'3 cot x + sin 7^ (2) 237. — Solve the equations (1) 236. — + n) cot 6=n cot 6, taken on AB so that shew that A — mcotB. are unequal values of satisfying the equation a tan d + b sec 6=1, find a and b in terms of a and sin a /3, and prove that + cos a + sin 3 + cos 8= —^ 1 239. If ?<„ = sin ^ + cos''^, prove «i 240. A ^ . +a^ that 3 building on a square base ABCD has the sides of the base AB and CU, parallel to the banks of a river. An observer standing on the bank t)f the river furthest from the building in the same straight line as I) A finds that the side subtends at his eye an angle of 45°, and after walking a yards along the AB l>ank he finds that Prove that the 1>A lengtli of subtends the angle whose sine each side of the base http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight in yards J is is - -- *. 2 408/467 8/12/2019 Elementary Trigonometry by Hall and Knight MISCELLANEOUS EXAMPLES Prove the identities 241. (I) . (cosec^-sinyl)(sec^-cos^) = (tan /l+cot^)-i; tan 6 , ^^^ cot 6 (1 -ftan^' 6f + 1 TcoW? = 2 (1 If sin 4(9 eos ^ 243. Prove that 2m% 1 tan-i = + sin -- , , . ' ^ ^^- 5^ bS — cos — 1 242. , find +tan-i -/^ o;i,<i value of 6. 2i/iV 2«(/ ^,, .„ , = tan-i ,-^-Wt, M=mp-nq, Nr=np + mq. wlêre 244. 365 K. In any triangle, prove that tan- tan- tan^ {a-h){a-c) 245. If j-j, \b-c){h-a)^ {c-a){c-h) be the radii of the three escribed >-3 ?•.,, A' circles, and shew that the 246. The triangle must be right-angled. sides a .triangle are 237 and 158, and the contained angle is 58° 40'of 3-9 Find by the aid of Tables the value of the base, without previously determining the t)tlier angles. 247. If tan {A + 5) = 3 tan sin {2 A 248. A, shew that + 25) + sin 2A = 2 sin 2B. Prove that 4 sin {0 - a) sin = 1 + (M >s (26 - 2vie) (me-n) cos {6-mff) - v.oH (20 - 2a) - cos (2m0 - 2a). http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 409/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 366 249. drawn from the angles A, B, C of on the opposite sides, and produced to Perpendiciilars are an acute-angled triangle meet the circumscribing circle respectively, shew that a 250. 90° If A if : those produced parts be a, /3, y + - + - = 2 (tan ^ + tan 5+ tan C). p and y B and are two angles, each positive, less than and such that 3sinM + 2sin2 5 = l, 3 sin 24 -2 sin 25=0, prove that 251. J^ + 25 = 90°. Prove that cot~i (tan 2x) (1 ,^, ^1—x , tan-i- (2) + cot-i - tan ( .1-1/ Sx) =x ; y—x tan-i—-^ = sm _i^--p—^ , • . j= In any triangle prove that 252. a sec ^=6-rc, where (Z) + c)sin^ = 2 V^ccos ^ Compass observations . are taken from a station to two points The bearing of distant respectively 1250 yards and 1575 yards. 7° 30' is 42° 15' other West of North, and that of the one point is the points by the aid East of North. Find the distance between of Tables. 253. Prove the identities ^ cos (1 cos (2) (x tan 254. (.S- - 0) 0+3/ cot If 2>S'= cos2 ,S'+ cos2 (2a - = cos^ a - sin^ (a - /3) a) A + B + C, >4) +cos2 (x cot a +i/ tan a) = (x +y)2 + 4xy cot2 2a. shew that (>S'- 5) + cos2 (,9 - C) =2+2 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight cos A cos B cos C. 410/467 8/12/2019 Elementary Trigonometry by Hall and Knight MISCELLANEOUS EXAMPLES If 255. 256. siuîa + sin~i(3 + sin~i'y = 7r, prove that 5 = 73° In a triangle a = 36, by the aid of Tables. 257. 3C7 K. 15', C=45° 30' ; find R and r If p be the radius of the circle inscribed in the pedal triangle, prove that p =^ ( 1 - cos2 A - cos2 B - cos2 C). A, B, C are the tops of posts at equal intervals by the side of a road ; t and i' are the tangents of the angles which AB and BC subtend at any point T is the tangent of the angle which the road makes with FB shew that 258. F ; : T~f 259. In any triangle prove that (cos5 + cos(7)(l 1 260. t' + cos A — 2 With the notation AI + 2cos4)_ b+c cos^ A a of Art. 219, prove that BI ££_ AI.^BI^'^CI,261. Prove that ^^ 262. '^+^^ 37n+^ ^ ;î = i- Find the relation between cot a cot /3 cot y — cot a, /3, and y a — COt j3 in order that — cot y should vanish. 263. U A+B + C=iT, tan A tan ,+ B ^.,_ tan +. C B tan C tan C tan A tan A tan B A + tan 5 + tan C- 2 (cot .4 + cot i? + cot C). tan = tan prove that . . http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 411/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 368 A man due North along a straight road observes that at a certain milestone two objects lie due N.E. and S.W. respectively, and that when he reaches the next milestone their directions have become S.S.E. and S.S.W. respectively. 264. travelling Find the distance between the two objects, and prove that the of their shortest distances from the road is exactly a mile. sum Prove that 265. tan 20° 266. + tan 40 + ^/3 tan If if sin 268. a-, b^, c- 269. + 6 cosec 2(9-8 cosec'' '20 = 0. B cos C cos A + cos- A = cos^ B A+B + C=0, 1 cos^ G, + 2 sin ^ sin C cos A + cos^ A = cos^ B + cos^ A , cot B, cot C C. are in a.p., shew that are also in a.p. If a, /3, -y are the angles of a triangle, prove that y-2a) + cos(|' + «-2/ii) + cos(|+/3-2y) 5^-2y-a 5a-2/3-7 =4 270. -\- prove that If in a triangle cot (f + cos <9 A -\-£+C= 180°, prove that 1—2 and = ^/3. Solve the equation cot3 267. 20° tan 40° If the cos sum ; ' 4 4-^ cos 5v-2a-/3 cos — 4 . of the pairs of radii of the escribed circles of a triangle taken in ordei- roiuid the triangle be denoted by and the corresponding difterences Ity c/j, d,^, d^, prove ,<tj, if.,, .s-^, that d^d.,d-i 271. If i;os,l=' , + dyi.^s.^ + <;/2*'3*'i + '^^3*r'*2 — ^• slifw that -4 .i2 sill sni - - http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight =1 '• 412/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight MISCELLANEOUS EXAMPLES 272. Prove that all 369 K. angles which satisfy the equation + 2tan(9 = l, tan2^ ± are included in the foi'nuila (8?i- 1) 5 7 where n zero or is any integer. 273. Prove (1) cos (2 tan-1 - ,. ^ ^ ' (2) 274. , tan-i ~ 5 Wsin 3 sin 2a -~ + 3 cos + , M tan-i ~\ — /tan a\ _, , ; tan-i 2a =n. —/ V -^ any triangle If in A fh \ c ^^^2=2Vc + 6' shew that the square described with one diagonal side of the triangle as equal to the rectangle contained by the other two is sides. 275. and C, = -0755470, log 9-7 = -9867717, 50° J having given log 1-19 Z sin 276. B Find Xsin 6= =50 , 119, « = 97. 70° = 9-9729858, Zsin70° l' = 9-9730318, = 9-8842540. Circles are inscribed the triangles D]Ê^t\, in JJ.yl'J.iF.,, Z>,, Z\, F^ are the points of contact of the circle Shew that if r^^, I'h, y^ be the radii of escribed to the side BC. l)rÊ.^F.^, where these circles 1 — '•« 277. I : I — : n —= /•<; Reduce ^ 1 A — tan — . ^ : 1 B — tan — . 4 4 to its simplest 278. If cos the sides are 1 — tan C -r 4 form tan-if.^^^.Vcot-1^ 1 : — .:rsin^/ \.%' C J +cosi5 = 4 sin^— in cos^ — smdJ' any triangle, shew that in a. p. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 413/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 414/467 , 8/12/2019 Elementary Trigonometry by Hall and Knight MISCELLANEOUS EXAMPLES 288. a , circle 371 K. Prove that the side of a regular heptagon inscribed in of radius unity is given by one of the roots of the equation and give the geometrical 289. signification of the other roots. If in a triangle the angle B is 4.5°, prove that (l+cot^)(l+cotC) = 2. 290. to the If twice the square sum equal that .1 sin''^ and that the + sin^ B + sin'^ C= 2, triangle is right-angled. If cos A = tan prove that 292. circle is of the squares on the sides of the inscribed triangle ABC, prove 291. on the diameter of a B, cos B = tan A = s,mB = sin C= 2 sin In any triangle shew that equation x^ - 25A-2 293. Shew that 294. The C, cos C= tan A sin 18°. a, b, c are the roots of the + (r2 + s^ + 4Rr) x - ARrs = 0. sin tt is a root of the equation stones from a circular field (radius /•) are collected Prove that the distance a labourer will have to travel with a wheelbarrow, which justs holds one heap, in bringing them together to one of into n heaps at regular intervals along the hedge. the heaps (supposing 295. Shew TT him to start from this heap) is 4rcot — that 2n Sn 4n- 5jr Qn 7n cos :— cos ,— cos VT cos q-= COS j-^ COS =— COS 15 15 15 15 15 15 15 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight /1\'^ \2j 415/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRV. 372 296. If ^v, y, are the jêrpendiculars z any point within a from drawn shew that triangle, to the sides x^-\-y^-\-^ is a minimum when _z _ X a _yb 2A a^ c + b^-\-c^' be the radii of the circles which touch each side and the adjacent two sides produced of a quadrilateral, prove that 297. If r„, Vh, r^, Vfi a c b â r, r^ -+-=- + d -. r^ If the diameters AA\ BE, 298. of the circum-circle in P, Q, It respectively, prove that cut the sides BC, CA, CC AB AP^BQ'^CR~R' 299. If a, the satisfy •' (3, y are angles,A unequal and eci nation ^ sin (a 300. / I TT ., + sm + 3) + sin (/3 -. ^ + c = 0, ' than irr, which prove that ^ + y) -f sin (y + a) = 0. Sliew that .,277 ,,'-iTr\ sec - + SCC- -— +sec/ \ cos ^ — —. less 7 -;:;- / I / ,, coscc TT , ., 4-cosec-' 7 / V http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight ^TT ..SttX / i -^ +cosec--^ ) = 192. J 416/467 8/12/2019 Elementary Trigonometry by Hall and Knight TABLES OF LOGARITHMS OF NUMBERS, ANTILOGARITHMS, NATURAL LOGARITHMIC SINES, COSINES, AND AND TANGENTS. 2(5 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 417/467 8/12/2019 374 Elementary Trigonometry by Hall and Knight ELEMENTARY TRTOONOMETRY. Logarithms. No. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 418/467 8/12/2019 / 1 1 /v Elementary Trigonometry by Hall and Knight / LOnARITHjrS. ^ 375 Logarithms. No. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 419/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 420/467 8/12/2019 Elementary Trigonometry ANTILOUAKITHMS. by Hall and Knight 377 Antilogarithms. ic http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 421/467 8/12/2019 Elementary Trigonometry by Hall and Knight 37a ELEMENTAKY TKIGONOMETKY. Natural Sines. to http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 422/467 8/12/2019 Elementary Trigonometry by Hall and Knight NATURAL SINES. 379 Natural Sines. be http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 423/467 8/12/2019 Elementary Trigonometry by Hall and Knight 380 ELEMENTAKV TRIGOXOMKTliV. Natural Cosines. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 424/467 8/12/2019 Elementary Trigonometry by Hall and Knight NATURAL COSINES. 381 Natural Cosines. ^ http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 425/467 8/12/2019 Elementary Trigonometry by Hall and Knight 382 ELEMENTARY TRIGONOMETRY. Natural Tangents. bi) http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 426/467 8/12/2019 Elementary Trigonometry by Hall and Knight NATURAL TANGENTS. 383 Natural Tangents. 6<3 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 427/467 8/12/2019 Elementary Trigonometry by Hall and Knight 384 ELEMENTAHY TKIUOXOJIETJIV. Logarithmic Sines. :'f. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 428/467 8/12/2019 Elementary Trigonometry by Hall and Knight I.OOARTTHMTC SINKS. 38.^ Logarithmic Sines. ._• http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 429/467 8/12/2019 386 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. Logarithmic Cosines. to http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 430/467 8/12/2019 Elementary Trigonometry by Hall and Knight LOGARITHM IC COSINES. 387 Logarithmic Cosines. to http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 431/467 8/12/2019 388 Elementary Trigonometry by Hall and Knight KT-RMRNTA RY T RTOOXOMKTRY. Logarithmic Tangents. til http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 432/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 433/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 434/467 8/12/2019 Elementary Trigonometry by Hall and Knight ANSWERS. I. Page 4. 1. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 435/467 8/12/2019 Elementary Trigonometry by Hall and Knight KLKMENTARY TRIGONOMETRY. 392 v'l + cotâ, - ^. -r m^ 14. 1 „ . H. 17. m- - , — 16. -i 1 r- 15. ^ 1 , -t; , • s. î— -s ^V-^9- IV. . cos ^sec-^-1 9. _ . Jl + cotâ p 13. ^ 24 cot a -^r— ,- 8. 24 ,^48 7 Page a. 3. 0. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 26. 4. 2*. 5. ^. 436/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 437/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. ^94 V. 1. 10^/3^17-32, 3. ^£^17-32 5. 22-56. 7. 20(3 9. 13-382, 36° 25'. Page b. Action., AD = + ^/3) = 94-61. VI. 277-12 1. 173-2 5. 22-5 8. 51 yds., 81 yds. 2. ft. ft.; 38-97 11. 273-2 14. 73-2 17. 1193 yds. a = 10v'2:=.14-ll, f^20. 2. ft., 8-66 22-89. 8. DC :^BD ^100. 60°. 6 30 64 ft, \/ 9, 86-6 yds ft. ft. 100 4. 50 7, 200 yds, ft. Each = 70-98 ft, y 42. 3 ft. 12, 4. 4. ft. 6. Page a. ft, 39. ft.; 10. 46-19 13, 5 miles. 16. 300 ft, ft. 277-12 yds. VI. Page b. 4 1. 565-6 yds.; 1131-2 yds. 2. 3-464 miles 3. 29 miles. 4. 10 miles per hour. 5. 10 miles 6. 16 miles 7. 9-656 miles. 8, 5-77 miles; 11-54 miles, 9. 295-1 knots. 10, 5-196 miles per hour 31 minutes past midnight, 12, 38-97 miles per hour. 11. ft, ; 24-14 miles. VI, Page c. 1. 36 yds. 1ft. 5, 24 yds. 6. 26° 34', 63° 26'. 8. 107 9. 244 ft. 2. 340 S. 25° W. 161-8 m. 3. 7. 10. 6 miles. 80° 668 yds. 26', 4. 14. 271m. 17. 441-5, 118-35 15, 7-9 mi, 16. 970 m. 18. N. 38°23'E, 19, 13-49 mi,, 24-12 mi. Page ft. 8'. 28° 15'. 13, ft. 586 11. 467-9 m., 784-7 m. a. 18 miles. 80° 26', 19° 12. VII. ; 48a. ft. ft. ; ; 54. 7ir '• 4- 6. -if^ 2Sw 72 ' 25' http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 438/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 439/467 8/12/2019 Elementary Trigonometry by Hall and Knight 396 ELEMENTARif TRIGONOMETKV. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 440/467 8/12/2019 Elementary Trigonometry by Hall and Knight 397 ANSWERS. X. Pagk a. 87. I. --^- 2. |. 3. J-6.- 4. -,/'2. 5. -^|. 6. 1. 7. -1. 8. -l- 9. 2. 10. -1. 11- -^- 12. -2. 13. -2. 14. -4j- 1^- ^' ^- ^^- ''' ''' 17. tan 18. -cos 19. -sec.^. 20. -cos J. 23. tan 24. - coscc 21. -tan^. 22. -cosO. 25. 1. 26. 2 sin J. 27. 1. .4.' .J. X. 0. II. -^- - -\- 1. 12. ^. • -^- 16. -^2- 21. ±30°, ±330°. 23. 120 , 300°, 30. 3. 13. -S./-2. '' ' cot- .4. 24 ^5 25- XI. 1 I. II. sin ^ cos cos A cos jy B cos C cos - cos sin 15. -v/2. ''• ^^• ^4 i> Jb- -1. Page 97. 36 100. 278 12 0; B 6' _^^ fi Pauk sin -4. 34. 16 65- cos -30°, -150°. 32. „ A -^. 135°, 315°, -45°, -225°. 3. cos 14. 24. . C- 5. 210°, 330°, b. ^. 2. I. -^. 22. a. 33 65' ^ ^- -1. 4. • -60°, -240^. 31. tf. »l. l 3. XL *• Page b. (^. -^; cos i? sin - sin A sin JJ cos C; 35. sin C- + cos A http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight sin cos A i> 1 4. C + sin ^ -^. C sin sin B sin C 441/467 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETRY. 398 tan 12. 1 - tan cot 13. 1. Ax4 tan B tan cot cot -B cot B tan C C + tan C tan A + tan ^ tan B B - tan G - tan A B cot O - cot A C + cot C cot tan - cot 5 A + cot ^ C B -1' - cot cot - 1. 4. 7. 10. 12. 14. 16. 18. 20. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 442/467 8/12/2019 Elementary Trigonometry by Hall and Knight 399 ANSWERiJ. XII. 1-2 cos 10. 1 14. sin 10a 18. sin 2^ =-96; cos 26. sin 2^ + Pagk f. .^. sin 6a ^ -8 ; + sin -la. cos 2^ 120 1. 60°. 5. 90°. 7. J = 30°, £ = 135°, = - -6 ^ ; . 15. 2-4936. Page 3. ^=:30°, 6. .4 15. 5=2^6, 4 = 75°, C=30°. 17. C=75°, a=c = 2^3 + 2. 19. C = 30°, a = 2, 24. 105°, 45°, 30°. 12. 7. 5 = 120°, = ^3 + 1. .-1 21. 2. 22. C = 90°, 30°; c = 2, 2. J5 = 60°, 120°; 4 = 75°, 15°; a = 3 + ^3, 3. 4 = 45°, 5 = 75°, b=;^3 + l; no 6. C=45°, 135°; 4 6. C=75°, 105°; 4 = 45°, 15°; 5 = 90°, = 105°, 15°; 9. C'= 72°. = ^/3 + l, c = V323. 6. 26. 32'. ^ , 1- 60°. 105°, 15°. 132. 120°; 8. 14, « Page b. 101° 9. = 105°, 5=60°, = 75°, C=60-'. 105°, 15°, 60°. 25. . 45°. 4. = ^5 + 1, 5 = 36°, a 16. or 90 12', C=80°. 14. 13. 8. 18. ti^ 50 28° 56'. 1. -4 10*. 16. 128. 8. XIII. 7. '28'. 27, = 75°, B = 45°, 6'=15°. ;^6. 6 -352; 69 ^ 63° 26'. a. 10. 11. n. 2^:::= -28. XIII. 2. 122.V. 1. 3-^/S. ambiguity. 4. Impossible. 9. Impossible. a=3+^3, 3-^3. a = 2;^3, 3-^3. 5 = 90°, 60°; 6 = 2^6, 3;^2. C= 72°, c = 4 ^5 + 2 ^5 no ambiguity. 105°; ; XIII. d. Page 136. 1. 72°, 72°, 36°; 2. 4 = 60°, 3. 4 = 105°, 5 = 15°, C= 60°. 5. C=60°, 120°; 4=90°, 30°; a = 100^3. 6. 18°, 126°. a= each side=v'5 9-3V3, 6 + l. = 3(^/6-^2), 8. 4. c = 3^2. 5 = 54°, C = 108°, 36°. No, for C=90°. 4 = 90°, 5 = 30°, http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 126°; (7=60°; 2c = a V3. 443/467 8/12/2019 Elementary Trigonometry by Hall and Knight 400 ELEMENTARY TRIGONOMETRY. MISCELLANEOUS EXAMPLES. cc,l. 2. 43. 9. A ^30°, £ = 75°, C=75°. 3. -1. 4. XIV. a. Page 6. D. (t^2, Pauk 138. B = 30°, C^lOo''. 145. 1. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 444/467 8/12/2019 Elementary Trigonometry by Hall and Knight ANSWERS. XV. 1. -616482.5. 2. b. -7928863. Page 3. 401 159. 1-21548.S8. 4. 62° 42' 31 . 30° 40' 23 . 6. 48° 45' 44 . 7. 8. 10-1317778. 9. 9-7530545. 10. 44° 17' 8 . II. 55° 30' 39 . 12. 9-6656561. 13. 10-1912872. 9-8440.554. 5. XV. c. Page 161. 1. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 445/467 8/12/2019 Elementary Trigonometry by Hall and Knight 402 ELEMENTARY TRTfiONOArETRY. 9. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 446/467 8/12/2019 Elementary Trigonometry by Hall and Knight ANRWKRP. =28° 20' 49 C = 39° 35' , 5. yl 7. (1) Not ambiguous, (2) ambiguous, 6 = 60-3893 (3) not ambiguous. for 11 . ^ =81° 45' 6. C = 90° XVI. 403 PaCxE 180. f. 2. 112° 12' 54 , 45° 53' 33 , 21° 53' 33 . 4. 4227-4815. 5 = 48° 11' 6. J = 105°38'57 8. 108°26'6 , 53°7'48 , 18°26'6 . 23 , 5 = 80° 46' 26 -5 12. 4-0249. 14. J = 42°0'14 15. 41° 24' 35 . 17. 889-25.54 19. 44-4878 B = 15°.S8'57 . ft. ft. = 108° 12' 26 C= 49° 27' 34 , 9. 126°22'; 96°27', or 19°3'. or 109° 59' 4 . 11. 70° 0' 56 , 13. 41° 45' 14 . 18. 72° 12' 59 , 47° 47' 1 . 20. ^ = 102° 24. 5 = 32° 15' 25. a 27. J? 28. ^=26° 24' 23 29. 53° 17' 55 , or 126° 42' 5 . 30. ^ = 3r39'33 , C=96°1'27 , 31. 6 32. i>' 33. Base = 2-44845 34. 90°, nearly. 36. ^=72° 56' 38 , , J5 = 98°55', C=53°35'4 . 49 , (7 = 44° 31' c 26. C = 147° 28' 37 , c , B = 118° , 3' , 22 . 72° 26' 26 , 22. a = 1180-525. 17 , = 33-5917, B = 4-2° = 18-7254. 37 , a . 17-1 or 3-68. = 32° 50' = 4028-38, .54 . 7. C' = 1°1'23 75° 48' 3. , = 20-9059, . B = 55° 56' 46 , C = 82°3'. 16. 4 = 60° 5' 34 , (7 = 29° 54' 26 , 23. = 75° .B . C= 63° 48' 33-5 , P = 99°54'23 ^ = 27° 29' 56 21. C = 73°23'54 5. 10. 58 . ft.; v4= 58° , 2' ; 1. 24' 43 , 2 , or 23 a 18' 25 , (7 = 2934 -124, a & = 3232-846. = 4389 -8. 6 = 642-756. = 878-753. = 2831-67. 53' 29 , or 104° 6' 31 ; ft., 31' 53 , c altitude 35. (1) ^ =60° = -713321 54' 19 , or 32° 41' 17 . ft. impossible; = 12-8255. 37. 6' (2) ambiguous; = 60° 13' -52 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight , f (3) 63-996. = 19-523977. 447/467 8/12/2019 Elementary Trigonometry by Hall and Knight KLKMENTARY TRTfiOXOAfKTRY. 40A XVI. 1. 108° 38'. 4. .4 Page g. 90°. 2. 30° 50', i= 183l,. 131° 15', 6. 7. 116° 28'. 9. .4 C C ==107°6'. 13. 14. .4 15. 5= 129° 29', 16. 5 = 49° 11. = 38° 25° 31'. 8. = 60°, C = 81°50'. 0=67° 22'. .J =53° 8', 5 = 59° 30', jB = 51°35', C = 20°59'. 12. ^ = 1°3', C=118°27'. 10. = 97° 10', J3 15-5', (7=70° 29', 31'. 5 = 29° 41-5'. 20. 21. 68-41. 22. 65-42. 25. 79-75. 26. a 28. (1 31-5', 5 = 34° 15', 30. = 214-2, a = 26-71, 32. 44° 17' or 135° 43'. 36. J =31° 40', 37. -^ =26° 12', 38. .4 = 102° 40. 5 = 75° 41. .4=42°, 42. 41° 24'. = 23° 3', 5 = 33° 17. /I = 71= 19. 5= 132° 15'. 23. in., = 5-499 in. 29. b = 3-841, 33. (7 4-200. ?; 31. or 155°7', 5 = 26° 16'. 20', r = 29°24'> 30', i = 36-49. = 4-328 or 105° 24', = 65°24' or (7=134°26' or <- 24. 1215. 27. 130-3. = 4-762. 57° 18'. = 36°18', c = 2918. 34°36', f = 133-2 or 82-22. 4° 12', c = 2-32-5 or 23-84. yl C = 96°2', a = 878-8. 5 = 118° 48', 6 = 8.50-7. 57', 5 c = 223-4. c = 99-68. 35. 12' 7', 6 5 = 74°36' 5 = 24° 53' 34. .4 5 = 37° 37-5', c = 19-49. (7 = 13° 31', a = 102-6. J =97° A = 24° 18. 8'. 17°55'. = 28° 24', 5 B = 44°30', J= ^ = 27° 40', 5 = 95° 27', C = 56°53'. 5. 41° 3. 5 = 42° 3'. 39. 41° 45'. or 104° 7', 4 = 60° 56' or 32° 40'. = 55° 58', C = 82°2'. 44. 5 = 99° 54-5', C = 32°.50-5', = 18-72. 5=1°1', C' = 147°29', 53° 17' or 126° 43'. = 4391. 45. 46. 12-81. 43. rt rt 47. XVII. 1. 146-4 6. .^ ft. 2. v''>-'^l*'ii»'l*'^- «S0^/3 Page a. = 1524ft. 7. 19-53. 185. 5. in(^l() + ^/2) http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight ,ihl(n-h){t. = 45-7()ft. 448/467 8/12/2019 Elementary Trigonometry by Hall and Knight 405 9. 14. loir- 10. '.tf 750v'<3-lHH7tl. 15. XVII. 1. 12. 30 2. ft. V500-200i7i5 == 12 1060oft. 3. r20v'6 = 294ft. 11. 50 ^120 + 30 Page c. 5. Height = 40 ;^6 = 98 v/6 ft. ; lOU It. 195. ^5^ = 408ft. 106 ft. distance = 40 v''14 ( + J2) ^ 206 ft. = 696 yds. XVII. Page d. 197. 1. 5 miles nearly. 2. Height^ 3. 200-1 4. Height = 394-4 5. 6. Height = 916-8 Height = 45-91 7. 11-55 or 25-97 miles per hour. 8. Height ft. = 159-2 ^fS)^V2.i^2it. 5. ft. 2. 10. + (\i 4 f t. XVII. 1. '2U0 Pagk VM. b. « v'2 •18v'6--117 ''ft- 12. ft- 19-4 yds. ft. ft.; distance= -9848 mile. ft. distance ft. ; ; distance = 99-17 = 215-5 ; ; = 102-9 yds. distance = 406-4 ft. distance ft. ft. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 449/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight ELEMENTAUY TRIGONOMETRY. i(.)6 XVIII. Page 210^^,7 b. 7-18875 1. 2(i-46 sq.ft. 2, 9-585 yds., 4. '216-23 sq. 0. 128-352 in. 7. 57-232 8. 63-09 sq. it. ft. XVIII. sq. yds. 6. 101-78 ft. ft. Page c. j,' 218. -<-' .i(^'-?) XVIII. 1. If , 2-|. 7071 ,sq. vds. 223. 4. Diagonals 65, 63 6. 2^77 + 6^11. XVIII. 2. Page d. Page e. /:' \ 5. + y ?^ z Expression = cot A +coti>+cot C. 4. 7? 6. 120. =15% = 45^\ 135°; (7=10.5°, 15°; c +i 135°; vl=--105°, 15 ; a 11. (' 12. 10 miles; 10 ^/-i^V^' miles. 24. (1) 90°--, (•J) 180° - 2A 25. K.\i)r('Ssion 29. llir. :$0'; T 90° -:|^, , 13. X F. 20, 21 Page 228. = ^G + ^2, JC>-^2. = 2^:i, 4^3-6. „ I r 90°-^- 180° - 21} 29 08-3 yds., 35 35 yds. 7. A area 1764. 225. MISCELLANEOUS EXAMPLES. 3. ; , = sin-(a-/3). 180° - 2C'. 28. 21-3 miles jjor liuur, 2)irs. 10'. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 450/467 8/12/2019 Elementary Trigonometry by Hall and Knight 3. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 451/467 . . 8/12/2019 Elementary Trigonometry by Hall and Knight KLEMENTARY TRIGONOMETRY. 408 12. nv, or rnr 14. inr + -, ±^ or 13. — r, > '2»7r, or 2inr 4 + -^ . . 2 [In some of the following examples, the equations have to be squared, so that extraneous solutions are introduced.] 15. 2|^ + J,or2«. + J. (2n+l)7r 10 19. mr + ^ 21. ^ 4 „ 23. (2« or /)7r 4 0:=«7r±-, 4 (2n+l)7r _,._^7r + 7r. 6 TT TT = ;(7r±-, ,„ (-l)»^^^. -J, orf + 2 ' , + l )7r 16. tf, — mr±-. D rf) = n7r±-3 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 452/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight ANSWERS. A 4. sin— = - 2 -= 2 cos 5. >,/l 2 sin -^= - ^/l 2 cos — = - 6. + sin 4 + \/l~s ^ ^; ^1 + sin A - + sin -4 - - ,y/l sin A. - \/l - sin A ; ^1 + sin il + ,/l - sin /t. 2sin —= + ^/l + sin4+ >/l-sin J; 2 cos -= + ^/l + sin ^ - A sm- = ., 9. (1) 2nv-- and 2nir+r; (3) 27nr 4: cos- .. + — and 14. (1) =^f2cos(d-j\; 15. (1) =-.sec2^; 4. 6' 7. 210 34 . 3'ards. 2. 5. 8. 1 -. 1 + 1W3-.503 and 2j(7r+— y/l - sin ^. =tan2 . (2) XXI. 1440 yards. — =2sin('e-^V (2) XX. 1. 2mr + +— — = a^/i + sin A + No; (2) ^15 .AS sm- = -,cos_=--. 8. 2H7r 10. 2 sin sin 4. A3 = 7. . - s/l „ 13. 409 . 14. b. a. Page Page 260. 267. 342f yards. 4'35 . StSS . 3. 22 yards. 6. 11 ft. 10. 50 ft. 15. = K-^ 200 1 m-u. 11 iu. '3 -491. 2 17 '^W3_3Q.7, http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 453/467 8/12/2019 Elementary Trigonometry by Hall and Knight KLE.MENTARY TRKiONOME 410 XXI. 1. 12 miles. 5. 204 8. 10560ft. ft. 2. 150 2 in. 11. 8. 14. 45° 54' 33 , 6. 9. 12. 44° Page b. 54' 33 . 7. 13. (1) cos a; 18° 26' 6 . 104 (2) ft. ft. 8 in. 2 in. = 26' 13 - sin . a. 27 . MISCELLANEOUS EXAMPLES. 3. 80 4. 610ft., 1^/110 minutes -1. 5' 271. 15 miles. 3. ft. TKV. 6. XXIIL H. Page 2S3. 35 miles or 13 miles per hour. a. Page 291. 1. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 454/467 . 8/12/2019 Elementary Trigonometry by Hall and Knight ANSWERS. XXIII. sm4n^ n sm 3 — —— sin ^^ -- —- sin —1— sin 0. 4 0. 6. cot 7. - <^ ^ tan-i(« 14. tau-i n {n + l)-^. + =a cos / /cos ^-i- jr-^ (cosa =a 4 sin ^ a 15. 22. 9. . 2 '' i 2 11. tan-'.r 13. tan ' +^+7 sin M , 2 / ?/ + siua), '- Page a. a-^ a+/3 — 2 13. cot 2 ^. - tan i -^^^ . n+1 {l)-j , (h + 1)} - ^. 1). XXIV. .T ~ ' ' 12. .T 2'' sm cosecaUau(7t + l)a - tana}. - 2 siu-(9-2 sin'-2—-. 2 3. '— ^ 10. 2. 294. n 4 sin - 8. Page b. n 4. 5. 411 ^ ^;„'^ = ?) = sm — -^ —+-^ a . sin / ,3 / 2 - cos (sin a -^ . 301. — a-B 2 / . a). ^ ^ cos sin — •—i- ?- + /3-7„„„^ + 7-a,_7 + a-/ _4cos(^—^':--)ricos(^^L-^+-j. (1) ((t2 + />2)j;2-26c:c+c2-a2 = 0; (2) (a2 + /v2)2 ,j.2 _ 2 (a2 - h ^) {2c' -a^- h^-) x + a* + Z/^ + 4c* - 2a262 -4«V-4i2,''^ = 0. [ Use cos 2a cos 2y3 = cos- (a - /j) - sin- (a + yS) http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight . 455/467 . . 8/12/2019 Elementary Trigonometry by Hall and Knight KLEMKNTARY TRIGONOMETRY. 412 XXV. 1. 2. 24. 3. 4. Page a. 4. 2. 8. 7. sJa^-2absma + bK ^/p'^ 10. + 2pq sin a + q-. 11. Maximum = 2 sin - 12. Maximum = sin- 13. Minimum = 2 tan 14. Minimum = 2 15. Maximum = -. 16. Minimum = ;y/3. 17. Minimum =-, 18. Minimum = 6. 19. Minimum =1. 20. Minimum = 1. 21. ^(a + c)i^-sJb-^ + (a-c)^. 26. k l{a^+h'^ 4 +c ); k-jia -o a- + = 2. fo 0- 2. cosec - 5 ^. 25. + b + c) XXV. 1. .1-'^ Page b. +. / == «^ + 324. ^ - 3 //- . 4 (' - {a''-h')'^=16ah. 1) 4. 7/ 7. rt.-6-(a2 10. -15 a- + 8. + 62)=:l. 8. ,r3 + ?/3 = a:*. 12. .r2 + w'2 = 2. = = 1. ^ 0- 16. a2^.j2 = 2r-. 22. — + 24. xy = (y 29. (rt 31. (« - h) y=a+ 20. or ?j, - x) t&n /<)(»i 25. 62^ - (« + b)-} - 2 4 a-). l. xhj^~ a^^=a^. 9. (x 13. _ ^2')i.2_ ^ (.j.2_ 30. x + yf+(x-y)^ = 2. â 2 2 a-^;,2 _ 2 cos a + M) = 2mH. (•- 2 (2 6. {x + yy^+ {x -yyi = 2a}K {a (?/2_ a. =a .rYi-.rV' = 2. 2 + 2. V2. 2Vm9. 318. = 2. 21. _ V2 + y2 ^2=l- 4^^_j.2y2_ 26. a + b = 2ah. + y-^l&a-. = iabcm. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 456/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 457/467 . 8/12/2019 . Elementary Trigonometry by Hall and Knight ELEMENTARY TRIGONOMETKY. 414 48. tan2^. 51. (1)0; 56. &j\°. 60. 50 70. -1. 75. loV^ft., 84. 4-14. 95. 27° 45' 44 . 98. 2-^3. 102. 45, 53; 107. 4cosec2i9. 110. 2-80103, ±^. 49. 53. -2. (2) 57. ^6 ft. 66. + ^3) -^. ft., 97. 99. 6' 59. 60 2-39794, + 15^/3 + y^ ^,. (2)2. 3-141. = 4^ 77. 8-10. 9-76143 2^=--—; tan-=--. d2o 104. -598626, .v'-^ ft. 3-73239, 2-60206, 108. 67. 74. sin 33-2 . (l)cot(7; = 86-6 yds. -38021, 89. 58° ^. 50,^/3 -^3, -^, 55. -sin2^. 73. 15(3 880(3+ ^3) = 4164-16 yds. 50. 1-3802113, 2 7 1-8239087. 120°. 20 106. — 49° 28' 32 . 9-69897, 9-849485. 112. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 458/467 . 8/12/2019 . . Elementary Trigonometry by Hall and Knight ANSWERS. 176. 5 = 105°, C = 45°, 178. ii = 81° 47' 12 C = 68° 12' 48 180. 10000 191. 2310 sq.ft.; 193. (1) «7r, 197. 134-19 ft. 212. (1)(2« + 1)^, 214. 20 231. —-5~ or 98° 12' 48 ; 55 ±16' ^^ H7r 2?t7r 10°. 256. 9-65146, 264. ;^/2 281. '^^ê- 246. 3 -g. 7nr+^-. 222. 8. 37-27919. ±.^. 233. TT ,-, (2)«7r4-3, «7r + 205-4. — Stt 252. a 262. d 1224-35 yards. + ^ + y = {2n + l)^ — inr. or 109° 59' 3 or 20° 0' 57 . + ;3 + 7 + 5) 283. : 4 ^, 4 266. B = 70° 0' 57 C = 59° 59' 3 CAMBRIDOE 206. (2) HIT, T or - 20-5309. miles. 4. 226-87. + (-l) j; TT 242. cos (a ft. miles nearly. (1)^, -y-^c^; 279. 70 ft., 204. 235. 275. 66 ft., 219. ? 7r = 13orll; 64° 31' 58 . 186. = 34 c or 51° 47' 12 . ft. ,,, a = J'2. ft. -2 415 + cos (a + (3 - 7 - 277. 5) + cos ^ = 45°, £ = H2i°, (a d. + 7 - /S - 5) + cos (a + 5 - /5 - 7) e= ^2-^2. PRINTED BY JOHN CLAY, M.A. AT THK ITNIVKR.SITY PRESS. http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 459/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 460/467 8/12/2019 Elementary Trigonometry by Hall and Knight MESSRS MACMILLAN AND By H. S. HALL, and M.A., SEVENTH EDITION, KNIGHT, R. S. FOR SCHOOLS. Globe 8vo. (bound in maroon-coloured cloth), Answers (bound in green-coloured cloth), 4s. M. KEY TO NEW EDITION. Crown ANSWERS TO EXAMPLES ALGEBEA. Fcap. 8vo. Sewed. HIGHER ALGEBRA. for Schools. B.A. Revised and Enlarged. ALGEBRA ELEMENTARY PUBLICATIONS. CO.'S 3«. 8vo. 6d. \\'îth 6d 8s. ELEMENTARY IN 1«. A. Sequel to Elementary Algebra Fourth Edition, revised and enlarged. Crown 8vo. 7s. 6d. 'We have no hesitation in saying that, SCHOOL GUARDIAN— best books that have been published on the subject. one of the it is KEY. Crown 8vo. 10s. 6d. ALGEBRA FOR BEGINNERS. With Answers. in our opinion, Globe 8vo. 2s. 2s. 6d. It possesses the systematic arrangement and the lucidity which have gained so much praise for the works previously written by the authors in collaboration. GUARDIAN— ALGEBRAICAL EXERCISES AND EXAMINATION PAPEES. revised With without Answers. or Third Edition, 2s. 6d. Globe Svo. can strongly recommend the volume to teachers seek- and enlarged. SCHOOLMASTER— We ing a well-arranged series of tests in algebra. ELEMENTARY TRIGONOMETRY. 4s. Globe 8vo. 6rf. EDUCATIONAL REVIEW— On the whole it is the best elementary treatise on Trigonometry we have seen. KEY. Crown Svo. 8s. 6d. ARITHMETICAL EXERCISES AND EXAMINATION PAPEES. LOGABITHMS Answers. Third With an Appendix containing Questions in MENSUEATION. With or without AND Edition, revised and enlarged. Globe Svo. 2s. 6d. CAMBRIDGE REVIEW— A\l the mathematical work these gentlemen have given to the public is of genuine worth, and these exercises are no exception to the rule. The addition of the logarithm and mensuration questions adds greatly to the value. ALGEBRAICAL EXAMPLES HaU and Knight's Supplementary to ALGEBEA FOE BEGINNEES and ELE- ALGEBEA (Chaps. MENTAEY M.A. With or without Answers. I.-XXVH.). Globe Svo. By H. S. Hall, 2s. A SHORT INTRODUCTION TO GRAPHICAL ALGEBEA. By H. KEY. Crown S. 8vo. EASY GRAPHS. Hall, M.A. Eevised Edition. Globe Svo. Is. 3s. 6c7. Crown MACMILLAN AND 8vo. Is. CO., Ltd., http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight LONDON. 461/467 8/12/2019 Elementary Trigonometry by Hall and Knight MESSRS MACMILLAN AND CO.'S PUBLICATIONS. MA, and A SCHOOL GEOMETRY, By H. HALL, S. H. F. STEVENS, M.A. based on the recommendations of the Mathematical Association, and on the recent report Crown 8vo. of the Cambridge Syndicate on Geometry. Farts I. and II.— Part I. Lines and Angles. Rectilineal Figures. Part II. Areas of Rectilineal Figures. 6d. Key. 3». &d. Is. Part I. Separately. Fart HI.— Circles. part of Containing the substance of Euclid Book Part Is. Book IV. 1». braical Formulae. Book 36—37. III. 6d. Containing the substance of Euclid Book III. 1—34, and Farts I., H., III. in one volume. Fart IV. — Squares and Rectangles. Farts XI. and IV. Containing 2s. 6d. Geometrical equivalents of Certain Algethe substance of Euclid Book II., and 6t/. one volume. Is. &d. Key. 6*. 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MACMILLAN AND CO., Ltd., http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight LONDON. 462/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 463/467 8/12/2019 Elementary Trigonometry by Hall and Knight /5 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight - ^ oJyt - }l\ 464/467 8/12/2019 Elementary Trigonometry by Hall and Knight A^^ http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 465/467 8/12/2019 Elementary Trigonometry by Hall and Knight http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 466/467 8/12/2019 Elementary Trigonometry by Hall and Knight UNIVERSITY OF IIIIIIIIIIIIIIIIIIIIIIIIIIHII B.C. ~ ~ LIBRARY ' ' 3 9424 03408 4498 http://slidepdf.com/reader/full/elementary-trigonometry-by-hall-and-knight 467/467

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