Gr 12 Maths BOOKWORK (an extract from Gr 12 Maths 2 in 1) All the proofs you need to know! Paper 1: Series (a maximum of 6 marks) page 1 Paper 2: Geometry (6 theorems) and Trigonometry (3 proofs) (a maximum of 12 marks) page 1 to 3 Be sure to study the geometry and trigonometry summaries on page 4 to 6. Calculator instructions page 7 BOOKWORK: PROOFS PAPER 1 PAPER 2 Circle Geometry Theorems Arithmetic and Geometric Series Sn , the sum of n terms 1 Arithmetic Series: Sn = a + (a + d) + (a + 2d) + . . . + Tn . . . n terms & Sn = Tn + (Tn – d) + (Tn – 2d) + . . . + a . . . n terms ∴ 2 Sn = (a + Tn) + (a + Tn) + (a + Tn) + . . . (a + Tn) The line segment drawn from the centre of a circle, perpendicular to a chord, bisects the chord. Given : ?O with OP ⊥ AB To prove : AP = PB Construction : Join OA and OB s Proof : In Δ OPA & OPB (1) OA = OB . . . radii (2) Pˆ 1 = Pˆ 2 (= 90º) . . . given (3) OP is common â ΔOAP ≡ ΔOBP . . . SøS â AP = PB, i.e. OP bisects chord AB . . . n terms ∴ 2 Sn = n (a + Tn) ∴ Sn = n (a + Tn) 2 Tn = a + (n – 1)d But ∴ Sn = n [ a + a + (n – 1)d ] 2 2 ∴ Sn = n [ 2a + (n – 1)d ] Sn = a + ar + ar % r ) ∴ r Sn = ar + ar 2 + ar + ar – : ∴ Sn – r Sn = a – ar 3 3 + . . . . . . + ar n–2 + ...... + ar + ar n –1 n–1 + ar ... n ˆ at the centre and APB ˆ at the circumference. ?O, arc AB subtending AOB P 12 x y Then Â = x . . . radii equal â Oˆ 1 = 2x . . . exterior ø of ΔAOP ... Similarly, if Pˆ 2 = y, then Oˆ 2 = 2y ˆ = 2x + 2y = 2(x + y) = 2 APB ˆ â AOB the 'middle bit' falls away. n B P ˆ = 2APB ˆ To prove : AOB Construction : Join PO and produce it to Q. Pˆ 1 = x Proof : Let Geometric Series: 2 1 2 A The angle which an arc of a circle subtends at the centre is double the angle it subtends at any point on the circumference. Given : 2 O O 1 2 x y A Q B n ∴ Sn (1 – r) = a(1 – r ) ∴ Sn = a(1 − r n ) 1− r . . . for r < 1 or × ( − 1) a(r n − 1) : Sn = ( − 1) r −1 3 . . . for r > 1 To prove : Â + Ĉ = 180º & B̂ + D̂ = 180º Construction : Join BO and DO. − (a − b) b − a a − b = = d − c c − d − (c − d) Proof : Â = Let Then â Ĉ = The derivation of the future and present value formulae, Fv and Pv, using the sum formula of a geometric sequence, must be understood, but is not examinable. See Topic 5 (Financial Maths) in the Gr 12 Maths 2 in 1 study guide. â Â + Ĉ = & â B̂ + D̂ = 1 x 2 O D 1 B . . . ø at centre = 2 % ø at circumference s 360º – 2x . . . ø about point O 1 (360º – 2x) = 180º – x . . . ø at centre = 2 % 2 ø at circumference 180º s 180º . . . sum of the ø of a quadrilateral = 360º ˆ = 2x O 1 â Oˆ 2 = Annuities: Copyright © The Answer x Given : ?O and cyclic quadrilateral ABCD Choose either depending on the value of r, whichever gives you a positive denominator. Notice that : A The opposite angles of a cyclic quadrilateral are supplementary. C 4 D ˆ = APB ˆ , A 1 To prove : ˆ = 90º DAC & ˆ DPA = 90º B 2 . . . diameter ⊥ tangent 2 . . . ø in semi-? 1 ∴ Aˆ 1 + Aˆ 2 = 90º and 6 1 P Construction : Draw diameter AD and join DP. Proof : We have x A ˆ ˆ A 2 = P1 s . . . ø in same seg. EC B Q E F B̂ = Ê & Ĉ = F̂ s In Δ DPQ & ABC . . . construction (1) DP = AB (2) DQ = AC . . . construction (3) D̂ = Â C But stage 2 : corresponding øs s corresponding ø equal DP = AB and . . . construction â AB = AC DE DF AD = AE DB EC stage 3 : parallel lines . . . proportion theorem DQ = AC A . . . SøS . . . given The â PQ || EF . . . focal â DP = DQ point DE DF stage 1 : congruency . . . given â ΔDPQ ≡ ΔABC E Given : ΔABC with DE || BC, D & E on AB & AC respectively. To prove : AB = AC = BC DE DF EF Proof : A line parallel to one side of a triangle divides the other two sides proportionally. DB C To prove : â 1̂ = B̂ = Ê D B ΔABC & ΔDEF with Â = D̂ A i.e. DE || BC ² AD = AE P 1 Given : Remember : Always give REASONS for each statement ! ˆ ∴ Aˆ 1 = Pˆ 2 ( = APB) The Proportion Theorem If two triangles are equiangular, then their sides are proportional and, therefore, they are similar. Construction : Mark P & Q on DE & DF such that DP = AB & DQ = AC C Pˆ 1 + Pˆ 2 = 90º But D A ˆ , Given : Tangent AC and chord AB subtending APB with C & P on opposite sides of AB. 5 The Similar Δs Theorem The angle between a tangent to a circle and a chord drawn from the point of contact is equal to the angle subtended by the chord in the alternate segment. stage 4 : proportions Similarly, by marking S and T on DE and EF such that Construction : Join DC & BE 1 AD . h Area of ΔADE = 2 = AD Proof : 1 Area of ΔDBE DB DB . h 2 1 AE . h′ 2 Similarly : Area of ΔADE = AE 1 Area of ΔEDC EC EC . h′ 2 h′ D B SE = AB and ET = BC, it can be proved that : h E AB BC = DE EF â AB = AC = BC DE C DF EF â ΔABC and ΔDEF are similar. s Similar Δ s But : ΔDBE = ΔEDC and : ΔADE is common ... Δ are similar if : A : they are equiangular, and B : their sides are proportional on the same base DE ; between || lines, DE & BC In this proof, we show that A B s ∴ The Δ are similar . . . Both conditions, A and B, apply â Area of ΔADE = Area of ΔADE Area of ΔDBE Area of ΔEDC AD AE â = DB EC The converse statement says : B A s ∴ The Δ are similar 2 Copyright © The Answer TRIGONOMETRY 3 The Cosine rule Â Area, Sine & Cosine Rules 1 A a2 = b2 + c2 – 2bc cos A b c D The Area rule A Ĉ acute Area of ΔABC = 1 ab sin C Construction : Draw AD ⊥ BC Proof : 1 Area of ΔABC = ah 2 But Construction : B ... h = sin C b 2 C D a 2 a = BD + h ∴ a ... . . . Pythagoras 2 2 = (c – AD) + h 2 2 2 = c – 2c . AD + 2 b 2 2 = b + c – 2c . AD In ΔADC : ∴ h = b sin C 2 = c – 2c . AD + AD + h B C a 2 . . . Pythagoras ... ... AD = cos A b ∴ AD = b cos A 1 ∴ Area of ΔABC = ab sin C 2 h Draw CD ⊥ BA 2 Proof : b h 2 in : acute : in : ∴ a2 = b2 + c2 – 2bc cos A 2 The Sine rule = sin C sin A = sin B a b Construction : Proof : B̂ acute Always construct the height from a vertex not involved in the formula : D i.e. To prove : b c h B Draw CD ⊥ AB In ΔADC : h = sin A b ∴ h = b sin A In ΔBDC : ... a C ∴ ... Copyright © The Answer sin A sin B = , draw a height from C, not A or B. a b To prove : a = b + c – 2bc cos A, draw a height from B or C, not A. 30º sin A sin B = a b Similarly, by drawing a perpendicular from B, one can prove : ∴ To prove : Remember : & : ∴ b sin A = a sin B ÷ ab) 1 ab sin C, draw a height from A or B, not C. 2 Area = 2 2 150º 2 II sin (180º – θ) = sin θ θ h = sin B a ∴ h = a sin B Equating A 180º – θ sin 150º = sin 30º cos 150º = – cos 30º cos (180º – θ) = – cos θ y x x + y = 180º sin x = sin y cos x = – cos y Formulae for Compound Angles The derivation of the formula : cos (A – B) = cos A cos B + sin A sin B will not be examined. (However, it is always good to understand how a formula arises ! ). There is an easy guide to this derivation in Topic 6 in the Gr 12 Maths 2 in 1 Study guide. You do need to know how to derive all the other formulae from this formula for cos(A – B). See examples of this, also found in Topic 6 in the Gr 12 Maths 2 in 1 Study guide. sin A sin C = a c sin A sin B sin C = = a c b 3 3 ways to prove that a quadrilateral is cyclic GROUPING OF CIRCLE GEOMETRY THEOREMS I The grey arrows indicate how various theorems are used to prove subsequent ones. The 'Centre' group We use 3 converse theorem statements to prove that a quadrilateral is cyclic. converse of 'ø's in same segment' theorem Prove that one side subtends equal angles at the two other points (on the same side). x 2x diameter 2x x x Equal radii ! II The 'No Centre' group Equal chords ! x1 z1 y2 B Prove that a pair of opposite angles is supplementary. Equal chords subtend equal angles and, vice versa, equal angles are subtended by equal chords. Either prove : Â + Ĉ = 180º A or B̂ + D̂ = 180º D C converse of 'exterior ø of c.q.' theorem Prove that an exterior angle of a quadrilateral is equal to the interior opposite angle. Either prove : 5 2 7 1̂ = 2̂ 4 8 3 6 1 x There are 3 ways to prove that a quad. is a cyclic quad. Either prove : x1 = x2 or y1 = y2 or z1 = z2 or p1 = p2 p1 x 2 z2 p2 converse of 'ø's of c.q.' theorem III The 'Cyclic Quadrilateral' group y y1 or 3̂ = 4̂ or 5̂ = 6̂ or 7̂ = 8̂ 2 ways to prove that a line is a tangent to a ? We use 2 converse theorem statements to prove that a line is a tangent to a ?. x + y = 180º converse of 'tangent ⊥ radius' theorem IV The 'Tangent' group O Prove that the line is perpendicular to a radius at a point where the radius meets the circumference. Prove that OP ⊥ AB ; then AB will be a tangent A P B converse of 'tan-chord' theorem Equal tangents ! Prove that when the line is drawn through the end point of a chord, it makes with the chord an angle equal to an angle in the alternate segment. There are 2 ways to prove that a line is a tangent to a ?. 4 A y x B Prove that x = y; then AB will be a tangent Note : It is a good idea to draw a feint circle (as shown) to see this converse theorem clearly. Copyright © The Answer QUADRILATERALS - pathways of definitions, areas and properties - A Summary A Rectangle All you need to know! 'Any' Quadrilateral f a A Trapezium b m A || with one right ø 2 h Area = ℓ % b 1 d DIAGONALS are EQUAL a s Sum of the ø of any quadrilateral = 360º The Square DEFINITION : b c e A Parallelogram DEFINITION : Quadrilateral with 1 pair of opposite sides || DEFINITION : Quadrilateral with 2 pairs opposite sides || the 'ultimate' quadrilateral ! Area = s 2 Area = base % height Area = Δ 1 + Δ 2 P = 1 ah + 1 bh 2 2 Sum of the interior angles = (a + b + c) + (d + e + f) s = 2 % 180º . . . (2 Δ ) = 360º Copyright © The Answer D B || ABCD = ABCQ + ΔQCD rect. PBCQ = ABCQ + ΔPBA where ΔQCD ≡ ΔPBA Properties : It's all been said 'before' ! Since a square is a rectangle, a rhombus, a parallelogram, a kite, . . . ALL the properties of these quadrilaterals apply. C m . . . SS90º m â || ABCD = rect. PBCQ (in area) = BC % QC A Kite a 2 a 2 b See how the properties accumulate as we move from left to right. i.e. the first quad has no special properties and each successive quadrilateral has all preceding properties. Q A Rhombus = 1 (a + b) . h 2 'Half the sum of the || sides % the distance between them.' The arrows indicate various ‘ROUTES’ from ‘any’ quadrilateral to the square (the ‘ultimate quadrilateral’). These routes, which combine logic and fact, are essential to use when proving specific types of quadrilaterals. A Properties : 2 pairs opposite sides equal 2 pairs opposite angles equal & DIAGONALS BISECT ONE ANOTHER Area = 1 product of diagonals (as for a kite) 2 THE DIAGONALS • bisect one another PERPENDICULARLY • bisect the angles of the rhombus • bisect the area of the rhombus Given diagonals a and b s Area = 2Δ = 2 ⎛⎜ 1 b . a ⎞⎟ = ab 2⎠ m A || with one pair of adjacent sides equal or = base % height (as for a parallelogram) DEFINITION : Quadrilateral with 2 pairs of consecutive sides equal ⎝2 DEFINITION : 2 'Half the product of the diagonals' THE DIAGONALS • cut perpendicularly • the LONG DIAGONAL bisects: the short diagonal, the opposite angles and the area of the kite 5 Note : x y y x x x y y s 2x + 2y = 180º . . . ø of Δ, or ² x + y = 90º co-int. øs suppl. Quadrilaterals play a prominent role in both Euclidian and Analytical Geometry right through to Grade 12 ! 2 SUMMARY : TRIGONOMETRY (Gr. 12) ' SPECIAL LS ' THE RATIOS sin 30E ' & their ' DEFINITIONS : The ANSWER SERIES y r O H x r A H y x 3 1 2 tan 45E ' 1 2 1 and cos 60° ' 2 O A tan θ ' y x & THEIR “FAMILIES” : ' SIGNS : ' GENERAL FORMS ' CRITICAL VALUES : any ratio ' 90E - θ (an acute angle) (see denominator ! ) ' GRAPHS : ± that SAME ratio of ' ' θ sin(90E + θ) cos(90E - θ) = sin θ cos(90E + θ) = - sin θ AREA ' SINE rule 1 ab sin C 2 1 2 And in RIGHT-ANGLED Δ S : 2 bh as well 2 ' cos(A ± B) = cos A cos B 6 COSINE rule ' c 2 ' a 2 % b 2 & 2ab cos C ' sin 2x = 2 sin x cos x ' sin(A ± B) = sin A cos B ± cos A sin B & cos θ = 1 - sin θ sin A sin B sin C ' ' a b c ' use regular trig. ratios & the theorem of Pythagoras !!! ' COMPOUND LS ˆ sin2 θ = 1 - cos2 θ = cos θ For non-right angled Δ S : But remember : area of ∆= sin 2 θ % cos 2 θ ' 1 90E + θ (an obtuse angle) sin(90E - θ) = cos θ ' SOLUTION OF ΔS sin θ tan θ ' cos θ = ' CO-RATIOS (sine and cosine) ' MINIMUM & MAXIMUM VALUES : (sin θ & cos θ) ' MUTUAL RELATIONSHIPS 180° - θ 180° + θ 360° - θ -θ K sin A sin B ' cos 2x = cos2 x - sin2 x 2 = 1 - 2 sin x 2 = 2 cos x - 1 Copyright © The Answer CALCULATOR INSTRUCTIONS MEAN AND STANDARD DEVIATION REGRESSION & CORRELATION There are 3 phases for each calculator procedure : The equation of the regression line STEP 1 : how to get there Brief instructions for the CASIO fx - 82ES STEP 2 : how to enter the data STEP 3 : how to find the mean and the standard deviation. STEP 1 : Press/Select : MODE ; 2 (STAT) ; 2 (A + Bx ) (and, in the next column, A, B and r) Grouped Data/Frequency Tables Ungrouped Data Casio fx-82ES Enter the x and y values, each followed by =. Casio fx-82ES You'll see : [MODE] [2 : STAT] [1 : 1 - VAR] Enter each value, followed by [=] after the last value : [=] [AC] 1 2 3 X … … … FREQ automatically entered as 1 , , [SHIFT] [STAT] [5 : VAR] [3 : x σ n] [=] If this is not done, the mean and standard deviation will be added as entries in the data table. (These would then need to be deleted.) Enter each value, followed by [M+] To find the mean : [RCL] [ x ] To find the S.D. : [RCL] [σ x ] FREQ … … … Press AC STEP 3 : Press SHIFT STAT (on '1') ; 7 (Reg) ; then : For A, press 1 = or For B, press 2 = To find B after A, press AC and then do the whole of STEP 3. to move to the top of the right-hand column. Just like for A and B in the regression function : Brief instructions for the CASIO fx - 82ES To find the mean : [SHIFT] [STAT] [5 : VAR] [2 : x ] [=] To find the S.D. : You'll see : Stat Type in the correct frequencies followed by [=] each , time; after the last frequency : [=] [AC] 0. Sharp EL-531WH You'll see : [MODE] [1 : STAT] - to get into stats mode . . . then [0 : SD] Enter midpoint of each interval (or each score if provided), followed by [STO]. Then enter the frequency followed by [M+]. (Enter all the data like this.) To find the S.D. : STEP 1 : Press/Select : MODE ; 2 (STAT) ; 2 (A + Bx ) [SHIFT] [STAT] [5 : VAR] [3 : x σ n] [=] To find the mean : [RCL] [ x ] Copyright © The Answer 1 2 3 X … … … The correlation coefficient It is essential to use [AC] at the end of entering all the data. [MODE] [1 : STAT] - to get into stats mode . . . then [0 : SD] [MODE] [2 : STAT] [1 : 1 - VAR] [SHIFT] [SETUP] Scroll down (use arrow) [3 : STAT] [1 : ON] Use Note for Casio fx-82ES Sharp EL-531WH You'll see : Enter the midpoint of each interval (or each score if provided) followed by [=], then . . . After the last value : [=] - and not [AC] - then . . . To find the mean : [SHIFT] [STAT] [5 : VAR] [2 : x ] [=] To find the S.D. : STEP 2 : [RCL] [σ x ] 7 Stat 0. STEP 2 : Enter the x and y values, each followed by =. Press AC STEP 3 : Press SHIFT STAT (on '1') ; 7 (Reg) ; then : But, now : For r press 3 = If you have found A and B first, press AC before going back to the beginning of STEP 3.