Exam 6: Analytic Trigonometry

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Find the exact value of the expression. - cos⁡(sin⁡−145)\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⁡(tan⁡−1u−tan⁡−1v)\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)=cos⁡θ−1f ( x ) = \cos \theta - 1 . Solve f(x)=0f ( x ) = 0 .

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

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Solve the equation on the interval 0≤θ<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. - sin⁡θ=−2129,3π2<θ<2π\sin \theta = - \frac { 21 } { 29 } , \frac { 3 \pi } { 2 } < \theta < 2 \pi \quad Find cos⁡θ2\cos \frac { \theta } { 2 } .

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

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

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

(Short Answer)
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Use the Half-angle Formulas to find the exact value of the trigonometric function. - tan⁡165∘\tan 165 ^ { \circ }

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

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

(Multiple Choice)
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Find the exact value under the given conditions. - cos⁡α=13,0<α<π2;sin⁡β=−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)=2cos⁡θ+1f ( x ) = 2 \cos \theta + 1 . Solve f(x)=0f ( x ) = 0 .

(Multiple Choice)
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Solve the problem. -If sin⁡θ=−45\sin \theta = - \frac { 4 } { 5 } , and θ\theta terminates in quadrant IV, then find sin⁡2θ\sin 2 \theta .

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Find the exact value of the expression. - cos⁡−132\cos ^ { - 1 } \frac { \sqrt { 3 } } { 2 }

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Simplify the expression. - 4sin⁡2xcos⁡2x4 \sin 2 x \cos 2 x

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Establish the identity. - tan⁡(x−π4)=tan⁡x−11+tan⁡x\tan \left( x - \frac { \pi } { 4 } \right) = \frac { \tan x - 1 } { 1 + \tan x }

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Solve the problem. -Multiply and simplify: (tan⁡θ+1)(tan⁡θ+1)−sec⁡2θtan⁡θ\frac { ( \tan \theta + 1 ) ( \tan \theta + 1 ) - \sec ^ { 2 } \theta } { \tan \theta }

(Short Answer)
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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 ) } = ?

(Multiple Choice)
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