TrigonometryFinalExamGuide

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Trigonometry Final Exam Study Guide
General Study Tips:
•rework all tests and take-home quizzes, example problems from class, and tests in online test archive
•memorize formulas and names of formulas, and the cases to which you apply them
Topics to Know:
Algebra
•manipulate fractions, including complex fractions
•rationalize denominators
•factor
•apply the square root theorem, zero product property, and quadratic formula
Basic Trig Topics (Ch. 5):
• evaluate without a calculator all six trig functions of any angle in radians or degrees having a 0, 30, 45, 60, or 90 reference angle
•evaluate without a calculator all six trig functions of any angle, given one of the six trig function values and the quadrant
•apply concepts of cofunction and like reference angles to evaluate trig functions given angles that are complements or have the
same reference angle but are in different quadrants
•convert between radians and degrees
•angles of elevation and depression
•linear speed, angular speed, and arc length
Graphing:
•graph all 6 basic trig functions
• find amplitude and period
•rewrite all equations by factoring to find horizontal shift (-c/b)
•apply horizontal and vertical shifts
•graph sum and difference of basic trig and linear functions
Identities (Ch. 6):
•evaluate trig functions of a given angle using sum, difference, half, and double angle identities
•simplify expressions using identities
•evaluate double, half, sum, and difference identities of an angle, given trig functions of one or more original angles and the
quadrant(s)
•evaluate inverse trig functions and compositions of inverse functions
•solve trig equations for both all real values of the variable and for all values of the variable in a given interval
•prove trigonometric identities
Solving Triangles (Ch. 7)
•identify type of triangle (SSS,SAS,SSA,ASA,AAS)
•solve any triangle given at least one side and two other measures
•solve word problems involving the Law of Sines and Law of Cosines
•find the area of any triangle
Vectors (Ch. 7)
•perform arithmetic operations with vectors (scalar multiplication and vector addition/subtraction)
•find the magnitude and direction angle of any vector
•write vectors in component form and in terms of ๐‘–โƒ— and ๐‘—โƒ—
•find a unit vector in the direction of any vector
•find the dot product of two given vectors
•find the angle between two vectors
•solve word problems involving a mass on an incline
•solve word problems involving heading of boats and airplanes
Trigonometric Form of Complex Numbers (Ch. 7)
•convert between standard form and trigonometric form
•determine the modulus and argument of a given complex number
•multiply and divide complex numbers in trigonometric form
Trig Functions of an Acute Angle
๐‘ ๐‘–๐‘‘๐‘’ ๐‘œ๐‘๐‘๐‘œ๐‘ ๐‘–๐‘ก๐‘’ ๐œƒ
sin ๐œƒ =
cos ๐œƒ =
tan ๐œƒ =
โ„Ž๐‘ฆ๐‘๐‘œ๐‘ก๐‘’๐‘›๐‘ข๐‘ ๐‘’
๐‘ ๐‘–๐‘‘๐‘’ ๐‘Ž๐‘‘๐‘—๐‘Ž๐‘๐‘’๐‘›๐‘ก ๐‘ก๐‘œ ๐œƒ
โ„Ž๐‘ฆ๐‘๐‘œ๐‘ก๐‘’๐‘›๐‘ข๐‘ ๐‘’
๐‘ ๐‘–๐‘‘๐‘’ ๐‘œ๐‘๐‘๐‘œ๐‘ ๐‘–๐‘ก๐‘’ ๐œƒ
๐‘ ๐‘–๐‘‘๐‘’ ๐‘Ž๐‘‘๐‘—๐‘Ž๐‘๐‘’๐‘›๐‘ก ๐‘ก๐‘œ ๐œƒ
csc ๐œƒ =
sec ๐œƒ =
cot ๐œƒ =
โ„Ž๐‘ฆ๐‘๐‘œ๐‘ก๐‘’๐‘›๐‘ข๐‘ ๐‘’
๐‘ ๐‘–๐‘‘๐‘’ ๐‘œ๐‘๐‘๐‘œ๐‘ ๐‘–๐‘ก๐‘’ ๐œƒ
โ„Ž๐‘ฆ๐‘๐‘œ๐‘ก๐‘’๐‘›๐‘ข๐‘ ๐‘’
๐‘ ๐‘–๐‘‘๐‘’ ๐‘Ž๐‘‘๐‘—๐‘Ž๐‘๐‘’๐‘›๐‘ก ๐‘ก๐‘œ ๐œƒ
๐‘ ๐‘–๐‘‘๐‘’ ๐‘Ž๐‘‘๐‘—๐‘Ž๐‘๐‘’๐‘›๐‘ก ๐‘ก๐‘œ ๐œƒ
๐‘ ๐‘–๐‘‘๐‘’ ๐‘œ๐‘๐‘๐‘œ๐‘ ๐‘–๐‘ก๐‘’ ๐œƒ
Converting Between Degree & Radian Measure
To convert from degree to radian measure, multiply by
To convert from radian to degree measure, multiply by
๐œ‹
180°
180°
๐œ‹
Arc Length and Angular Speed
Variables
๐‘  = ๐‘‘๐‘–๐‘ ๐‘ก๐‘Ž๐‘›๐‘๐‘’ ๐‘ก๐‘Ÿ๐‘Ž๐‘ฃ๐‘’๐‘™๐‘’๐‘‘ or ๐‘Ž๐‘Ÿ๐‘ ๐‘™๐‘’๐‘›๐‘”๐‘กโ„Ž (๐‘–๐‘›๐‘โ„Ž๐‘’๐‘ , ๐‘˜๐‘–๐‘™๐‘œ๐‘š๐‘’๐‘ก๐‘’๐‘Ÿ๐‘ , ๐‘’๐‘ก๐‘)
๐‘ก = ๐‘ก๐‘–๐‘š๐‘’ (๐‘ ๐‘’๐‘๐‘œ๐‘›๐‘‘๐‘ , ๐‘š๐‘–๐‘›๐‘ข๐‘ก๐‘’๐‘ , โ„Ž๐‘œ๐‘ข๐‘Ÿ๐‘ , ๐‘‘๐‘Ž๐‘ฆ๐‘ , ๐‘’๐‘ก๐‘)
๐œƒ = ๐‘Ž๐‘š๐‘œ๐‘ข๐‘›๐‘ก ๐‘œ๐‘“ ๐‘Ÿ๐‘œ๐‘ก๐‘Ž๐‘ก๐‘–๐‘œ๐‘› or ๐‘–๐‘›๐‘๐‘™๐‘ข๐‘‘๐‘’๐‘‘ ๐‘Ž๐‘›๐‘”๐‘™๐‘’ (๐‘‘๐‘’๐‘”๐‘Ÿ๐‘’๐‘’๐‘ , ๐‘Ÿ๐‘Ž๐‘‘๐‘–๐‘Ž๐‘›๐‘ , ๐‘Ÿ๐‘œ๐‘ก๐‘Ž๐‘ก๐‘–๐‘œ๐‘›๐‘ , ๐‘Ÿ๐‘’๐‘ฃ๐‘œ๐‘™๐‘ข๐‘ก๐‘–๐‘œ๐‘›๐‘ , ๐‘’๐‘ก๐‘)
๐‘Ÿ = ๐‘Ÿ๐‘Ž๐‘‘๐‘–๐‘ข๐‘  ๐‘œ๐‘Ÿ ๐‘‘๐‘–๐‘ ๐‘ก๐‘Ž๐‘›๐‘๐‘’ ๐‘“๐‘Ÿ๐‘œ๐‘š ๐‘กโ„Ž๐‘’ ๐‘๐‘’๐‘›๐‘ก๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘Ÿ๐‘œ๐‘ก๐‘Ž๐‘ก๐‘–๐‘œ๐‘› (๐‘๐‘’๐‘›๐‘ก๐‘–๐‘š๐‘’๐‘ก๐‘’๐‘Ÿ๐‘ , ๐‘–๐‘›๐‘โ„Ž๐‘’๐‘ , ๐‘’๐‘ก๐‘)
๐‘‘๐‘–๐‘ ๐‘ก๐‘Ž๐‘›๐‘๐‘’
v = linear speed =
๐‘ก๐‘–๐‘š๐‘’
๐‘Ž๐‘š๐‘œ๐‘ข๐‘›๐‘ก ๐‘œ๐‘“ ๐‘Ÿ๐‘œ๐‘ก๐‘Ž๐‘ก๐‘–๐‘œ๐‘›
๐œ” = ๐‘Ž๐‘›๐‘”๐‘ข๐‘™๐‘Ž๐‘Ÿ ๐‘ ๐‘๐‘’๐‘’๐‘‘ =
๐‘ก๐‘–๐‘š๐‘’
Formulas
๐‘ 
θ
๐‘  = ๐‘Ÿ๐œƒ ,
๐‘ฃ= ,
ω= ,
๐‘ฃ = ๐‘Ÿω
๐‘ก
t
Dimensional analysis conversion factors
5280 ๐‘“๐‘ก 12 ๐‘–๐‘› 2๐œ‹
๐œ‹
60 ๐‘š๐‘–๐‘› 60 ๐‘ ๐‘’๐‘
,
,
,
,
,
, ๐‘Ž๐‘›๐‘‘ ๐‘กโ„Ž๐‘’๐‘–๐‘Ÿ ๐‘Ÿ๐‘’๐‘๐‘–๐‘๐‘Ÿ๐‘œ๐‘๐‘Ž๐‘™๐‘ 
1 ๐‘š๐‘–
1 ๐‘“๐‘ก 1 ๐‘Ÿ๐‘’๐‘ฃ 180° 1 โ„Ž๐‘Ÿ
1 ๐‘š๐‘–๐‘›
Example problem: A car travels at 60 miles per hour. Its wheels have a 24-inch diameter. What is the angular speed of a point on
the rim of a wheel in revolutions per minute?
Solution: ๐‘ฃ =
60 ๐‘š๐‘–
1 โ„Ž๐‘Ÿ
, = 12 ๐‘–๐‘› , ๐œ” = ?
Equation relating these variables: ๐‘ฃ = ๐‘Ÿ๐œ”
๐‘ฃ
1 60 ๐‘š๐‘–
1
1 โ„Ž๐‘Ÿ 5280 ๐‘“๐‘ก 12 ๐‘–๐‘› 1 ๐‘Ÿ๐‘’๐‘ฃ 2640 ๐‘Ÿ๐‘’๐‘ฃ
๐œ”= =๐‘ฃโˆ™ =
โˆ™
โˆ™
โˆ™
โˆ™
โˆ™
=
๐‘Ÿ
๐‘Ÿ
1 โ„Ž๐‘Ÿ 12 ๐‘–๐‘› 60 ๐‘š๐‘–๐‘› 1 ๐‘š๐‘–
1 ๐‘“๐‘ก
2๐œ‹
๐œ‹ ๐‘š๐‘–๐‘›
Reciprocal Identities
1
csc ๐‘ฅ =
sin ๐‘ฅ
1
sin ๐‘ฅ =
sec ๐‘ฅ =
cos ๐‘ฅ =
cot ๐‘ฅ =
csc ๐‘ฅ
1
cos ๐‘ฅ
1
sec ๐‘ฅ
1
tan ๐‘ฅ
1
tan ๐‘ฅ =
cot ๐‘ฅ
Ratio Identities
tan x =
cot ๐‘ฅ =
sinx
cos x
cos ๐‘ฅ
Odd-Even Identities
cos(−๐‘ฅ) = cos ๐‘ฅ ,
sin(−๐‘ฅ) = − sin ๐‘ฅ ,
tan(−๐‘ฅ) = − tan ๐‘ฅ
sec(−๐‘ฅ) = sec ๐‘ฅ ,
csc(−๐‘ฅ) = − csc ๐‘ฅ ,
cot(−๐‘ฅ) = − cot ๐‘ฅ
Sum and Difference Identities
sin(๐‘Ž + ๐‘) = sin ๐‘Ž cos ๐‘ + cos ๐‘Ž sin ๐‘
sin(๐‘Ž − ๐‘) = sin ๐‘Ž cos ๐‘ − cos ๐‘Ž sin ๐‘
cos(๐‘Ž + ๐‘) = cos ๐‘Ž cos ๐‘ − sin ๐‘Ž sin ๐‘
cos(๐‘Ž − ๐‘) = cos ๐‘Ž cos ๐‘ + sin ๐‘Ž sin ๐‘
tan(๐‘Ž + ๐‘) =
sin ๐‘ฅ
Pythagorean Identities
sin2 ๐‘ฅ + cos2 ๐‘ฅ = 1 ,
1 + cot 2 ๐‘ฅ = csc 2 ๐‘ฅ ,
tan2 ๐‘ฅ + 1 = sec 2 ๐‘ฅ
tan(๐‘Ž − ๐‘) =
tan ๐‘Ž+tan ๐‘
1−tan ๐‘Ž tan ๐‘
tan ๐‘Ž−tan ๐‘
1+tan ๐‘Ž tan ๐‘
Cofunction Identities
๐œ‹
sin ( − ๐‘ฅ) = cos ๐‘ฅ
2
๐œ‹
Half-Angle Identities
,
cos ( − ๐‘ฅ) = sin ๐‘ฅ
2
๐œ‹
tan ( − ๐‘ฅ) = cot ๐‘ฅ
2
๐œ‹
,
2
csc ( − ๐‘ฅ) = sec ๐‘ฅ
2
๐œ‹
,
sec ( − ๐‘ฅ) = csc ๐‘ฅ
2
Double-Angle Identities
sin 2๐‘ฅ = 2 sin ๐‘ฅ cos ๐‘ฅ
cos 2๐‘ฅ = cos2 ๐‘ฅ − sin2 ๐‘ฅ
cos 2๐‘ฅ = 2 cos2 ๐‘ฅ − 1
cos 2๐‘ฅ = 1 − 2 sin2 ๐‘ฅ
tan 2๐‘ฅ =
2 tan ๐‘ฅ
1−tan2 ๐‘ฅ
1−cos ๐‘ฅ
2
2
๐‘ฅ
1+cos ๐‘ฅ
cos = ±√
2
2
๐‘ฅ
1−cos ๐‘ฅ
2
๐‘ฅ
1+cos ๐‘ฅ
sin ๐‘ฅ
tan = ±√
cot ( − ๐‘ฅ) = tan ๐‘ฅ
๐œ‹
๐‘ฅ
sin = ±√
tan =
2
๐‘ฅ
1+cos ๐‘ฅ
1−cos ๐‘ฅ
2
sin ๐‘ฅ
tan =
Solving Triangles
Law of Sines
sin ๐ด
๐‘Ž
=
sin ๐ต
๐‘
=
sin ๐ถ
๐‘
or
๐‘Ž
sin ๐ด
=
๐‘
sin ๐ต
=
๐‘
sin ๐ถ
Law of Cosines
๐‘ 2 = ๐‘Ž2 + ๐‘ 2 − 2๐‘Ž๐‘ cos ๐ถ
๐‘ 2 = ๐‘Ž2 + ๐‘ 2 − 2๐‘Ž๐‘ cos ๐ต
๐‘Ž2 = ๐‘ 2 + ๐‘ 2 − 2๐‘๐‘ cos ๐ด
Area of a Triangle
1
1
1
๐พ = ๐‘๐‘ sin ๐ด = ๐‘Ž๐‘ sin ๐ต = ๐‘Ž๐‘ sin ๐ถ
2
2
2
Vectors
โƒ—โƒ—โƒ—โƒ—โƒ—โƒ— with ๐ด = (๐‘ฅ1 , ๐‘ฆ1 ) and ๐ถ = (๐‘ฅ2 , ๐‘ฆ2 ) is ๐ด๐ถ
โƒ—โƒ—โƒ—โƒ—โƒ—โƒ— = ⟨๐‘ฅ2 − ๐‘ฅ1 , ๐‘ฆ2 − ๐‘ฆ1 ⟩
The component form of ๐ด๐ถ
โƒ—โƒ—| = √๐’‚๐Ÿ + ๐’ƒ๐Ÿ
The magnitude of a vector ๐‘ฃโƒ— with component form ⟨๐‘Ž, ๐‘⟩ is |๐’—
๐‘
The reference angle ๐›ผ for the direction angle ๐œƒ of the vector ⟨๐‘Ž, ๐‘⟩ is given by ๐›ผ = |tan−1 |. Figure out
๐‘Ž
which quadrant this angle should be in and measure the angle counterclockwise from the positive x-axis.
The horizontal component of the vector ⟨๐‘Ž, ๐‘⟩ is ๐‘Ž = |๐‘ฃโƒ—| cos ๐œƒ
The vertical component of the vector ⟨๐‘Ž, ๐‘⟩ is ๐‘ = |๐‘ฃโƒ—| sin ๐œƒ
For a real number ๐‘˜ and a vector ๐‘ฃโƒ— = ⟨๐‘ฃ1 , ๐‘ฃ2 ⟩, the scalar product of ๐‘˜ and ๐‘ฃโƒ— is ๐‘˜๐‘ฃโƒ— = ๐‘˜⟨๐‘ฃ1 , ๐‘ฃ2 ⟩ = ⟨๐‘˜๐‘ฃ1 , ๐‘˜๐‘ฃ2 ⟩. The vector ๐‘˜๐‘ฃโƒ— is a
scalar multiple of the vector ๐‘ฃโƒ—.
โƒ—โƒ— ± ๐’—
โƒ—โƒ— = ⟨๐’–๐Ÿ ± ๐’—๐Ÿ , ๐’–๐Ÿ ± ๐’—๐Ÿ ⟩ .
Vector Addition/Subtraction: If ๐‘ข
โƒ—โƒ— = ⟨๐‘ข1 , ๐‘ข2 ⟩ and ๐‘ฃโƒ— = ⟨๐‘ฃ1 , ๐‘ฃ2 ⟩, then ๐’–
โƒ—โƒ—
๐’—
If ๐‘ฃโƒ— is a vector and ๐‘ฃโƒ— ≠ โƒ—0โƒ—, then |๐’—โƒ—โƒ—| is a unit vector (vector with magnitude 1) in the direction of ๐‘ฃโƒ—.
โƒ—โƒ— โˆ™ ๐’—
โƒ—โƒ— = ๐’–๐Ÿ ๐’—๐Ÿ + ๐’–๐Ÿ ๐’—๐Ÿ .
The dot product of two vectors ๐‘ข
โƒ—โƒ— = ⟨๐‘ข1 , ๐‘ข2 ⟩ and ๐‘ฃโƒ— = ⟨๐‘ฃ1 , ๐‘ฃ2 ⟩ is ๐’–
โƒ—โƒ—โƒ—โˆ™๐’—
โƒ—โƒ—
๐’–
If ๐œƒ is the angle between two nonzero vectors ๐‘ข
โƒ—โƒ— and ๐‘ฃโƒ—, then ๐œ๐จ๐ฌ ๐œฝ = |๐’–โƒ—โƒ—โƒ—||๐’—โƒ—โƒ—| .
Trigonometric Form of Complex Numbers
A complex number ๐‘ง = ๐‘Ž + ๐‘๐‘–, where ๐‘– = √−1 can be written in trigonometric form as ๐‘ง = ๐‘Ÿ(cos ๐œƒ + ๐‘– sin ๐œƒ) or ๐‘ง = ๐‘Ÿ cis ๐œƒ, where
๐‘Ÿ = √๐‘Ž2 + ๐‘ 2 is the modulus of ๐‘ง and direction angle ๐œƒ is referred to as the argument.
๐‘ง1 ๐‘ง2 = ๐‘Ÿ1 ๐‘Ÿ2 [cos(๐œƒ1 + ๐œƒ2 ) + ๐‘– sin(๐œƒ1 + ๐œƒ2 )] = ๐‘Ÿ1 ๐‘Ÿ2 cis ๐œƒ
๐‘ง1 ๐‘Ÿ1
๐‘Ÿ1
= [cos(๐œƒ1 − ๐œƒ2 ) + ๐‘– sin(๐œƒ1 − ๐œƒ2 )] = cis ๐œƒ
๐‘ง2 ๐‘Ÿ2
๐‘Ÿ2
๐‘ง ๐‘› = ๐‘Ÿ ๐‘› [cos(๐‘›๐œƒ) + ๐‘– sin(๐‘›๐œƒ)] = ๐‘Ÿ ๐‘› cis(๐‘›๐œƒ)
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