Exam 6: Analytic Trigonometry

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Find the exact value of the expression. - cos(sin145)\cos \left( \sin ^ { - 1 } \frac { 4 } { 5 } \right)

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Write the trigonometric expression as an algebraic expression containing u and v. - sin(tan1utan1v)\sin \left( \tan ^ { - 1 } \mathrm { u } - \tan ^ { - 1 } \mathrm { v } \right)

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Solve the equation on the interval [0, 2π). -Suppose f(x)=cosθ1f ( x ) = \cos \theta - 1 . Solve f(x)=0f ( x ) = 0 .

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Solve the problem. -If tanθ=724\tan \theta = \frac { 7 } { 24 } , and π<θ<3π2\pi < \theta < \frac { 3 \pi } { 2 } , then find tan2θ\tan 2 \theta .

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Solve the equation on the interval 0θ<2π0 \leq \theta < 2 \pi - csc(3θ)=0\csc ( 3 \theta ) = 0

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Use the information given about the angle θ, 0 ≤θ ≤ 2π, to find the exact value of the indicated trigonometric function. - sinθ=2129,3π2<θ<2π\sin \theta = - \frac { 21 } { 29 } , \frac { 3 \pi } { 2 } < \theta < 2 \pi \quad Find cosθ2\cos \frac { \theta } { 2 } .

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Complete the identity. - sin(α+β)cosβcos(α+β)sinβ=?\sin ( \alpha + \beta ) \cos \beta - \cos ( \alpha + \beta ) \sin \beta = ?

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Complete the identity. - sin(tan1v)=?\sin \left( \tan ^ { - 1 } v \right) = ?

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Use a calculator to solve the equation on the interval 0 0x<2π0 \leq x < 2 \pi . Round the answer to one decimal place if necessary. - ex=cosxe ^ { x } = \cos x

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Use the Half-angle Formulas to find the exact value of the trigonometric function. - tan165\tan 165 ^ { \circ }

(Multiple Choice)
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Find the exact value of the expression. Do not use a calculator. - sin1[sin(π5)]\sin ^ { - 1 } \left[ \sin \left( \frac { \pi } { 5 } \right) \right]

(Multiple Choice)
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Solve the equation on the interval 0θ<2π0 \leq \theta < 2 \pi - 2cos(2θ)=1\sqrt { 2 } \cos ( 2 \theta ) = 1

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Find the exact value under the given conditions. - cosα=13,0<α<π2;sinβ=12,π2<β<0\cos \alpha = \frac { 1 } { 3 } , 0 < \alpha < \frac { \pi } { 2 } ; \quad \sin \beta = - \frac { 1 } { 2 } , - \frac { \pi } { 2 } < \beta < 0 Find tan(α+β)\tan ( \alpha + \beta )

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Solve the equation on the interval [0, 2π). -Suppose f(x)=2cosθ+1f ( x ) = 2 \cos \theta + 1 . Solve f(x)=0f ( x ) = 0 .

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Solve the problem. -If sinθ=45\sin \theta = - \frac { 4 } { 5 } , and θ\theta terminates in quadrant IV, then find sin2θ\sin 2 \theta .

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Find the exact value of the expression. - cos132\cos ^ { - 1 } \frac { \sqrt { 3 } } { 2 }

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Simplify the expression. - 4sin2xcos2x4 \sin 2 x \cos 2 x

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Establish the identity. - tan(xπ4)=tanx11+tanx\tan \left( x - \frac { \pi } { 4 } \right) = \frac { \tan x - 1 } { 1 + \tan x }

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Solve the problem. -Multiply and simplify: (tanθ+1)(tanθ+1)sec2θtanθ\frac { ( \tan \theta + 1 ) ( \tan \theta + 1 ) - \sec ^ { 2 } \theta } { \tan \theta }

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Complete the identity. - cos(3θ)cos(9θ)cos(3θ)+cos(9θ)=?\frac { \cos ( 3 \theta ) - \cos ( 9 \theta ) } { \cos ( 3 \theta ) + \cos ( 9 \theta ) } = ?

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