Exam 6: Analytic Trigonometry

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Solve the equation. Give a general formula for all the solutions. - cos⁡(2θ)=22\cos ( 2 \theta ) = \frac { \sqrt { 2 } } { 2 }

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Find the exact value of the expression. - sin⁡(tan⁡−12)\sin \left( \tan ^ { - 1 } 2 \right)

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Complete the identity. - 1−cos⁡(2θ)+cos⁡(6θ)−cos⁡(8θ)=1 - \cos ( 2 \theta ) + \cos ( 6 \theta ) - \cos ( 8 \theta ) = ?

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Use a graphing utility to solve the equation on the interval 0 0∘≤x<360∘0 ^ { \circ } \leq x < 360 ^ { \circ } . Express the solution(s) rounded to one decimal place. - 2+13sin⁡x=14cos⁡2x2 + 13 \sin x = 14 \cos ^ { 2 } x

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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. - cos⁡θ=−35,π<θ<3π2\cos \theta = - \frac { 3 } { 5 } , \pi < \theta < \frac { 3 \pi } { 2 } \quad Find cos⁡θ2\cos \frac { \theta } { 2 } .

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Solve the problem. -The ground movement of an earthquake near a fault line is modeled by the equation d=Dtan⁡[π2(1−2MS)]d = D \tan \left[ \frac { \pi } { 2 } \left( 1 - \frac { 2 M } { S } \right) \right] where M is the horizontal movement (in meters) at a distance d (in kilometers) from the earthquake, D is the depth (also in kilometers) below the surface of the center of the earthquake, and S is the total horizontal displacement (also in meters) at the fault line. What is the horizontal movement 5 kilometers from an earthquake centered 3 kilometers below the surface with a total horizontal displacement of 4 meters? Round the answer to the nearest 0.01 meter.

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Find the exact value of the expression. - tan⁡−1[tan⁡(6π7)]\tan ^ { - 1 } \left[ \tan \left( \frac { 6 \pi } { 7 } \right) \right]

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Find the exact value of the expression. - sin⁡255∘cos⁡15∘−cos⁡255∘sin⁡15∘\sin 255 ^ { \circ } \cos 15 ^ { \circ } - \cos 255 ^ { \circ } \sin 15 ^ { \circ }

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Find the exact solution of the equation. - −sin⁡−1(4x)=π4- \sin ^ { - 1 } ( 4 x ) = \frac { \pi } { 4 }

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Solve the problem. -The seasonal variation in the length of daylight can be represented by a sine function. For example, the daily number of hours of daylight in a certain city in the U.S. can be given b h=414+53sin⁡2πx365h = \frac { 41 } { 4 } + \frac { 5 } { 3 } \sin \frac { 2 \pi x } { 365 } , where x is the number of days after March 21 ( disregarding leap year). On what day(s) will there be about 10 hours of daylight?

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Solve the problem. -If cos⁡θ=−513\cos \theta = - \frac { 5 } { 13 } , and π2<θ<π\frac { \pi } { 2 } < \theta < \pi , then find tan⁡2θ\tan 2 \theta .

(Multiple Choice)
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Find the inverse function f−1\mathrm { f } ^ { - 1 } of the function f. - f(x)=5tan⁡(10x−9)f ( x ) = 5 \tan ( 10 x - 9 )

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Write the trigonometric expression as an algebraic expression in u. - cos⁡(cot⁡−1u)\cos \left( \cot ^ { - 1 } \mathrm { u } \right)

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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. - sec⁡θ=4,0<θ<π2\sec \theta = 4 , \quad 0 < \theta < \frac { \pi } { 2 } Find cos⁡θ2\cos \frac { \theta } { 2 } .

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Complete the identity. - sec⁡θ−1sec⁡θ=\sec \theta - \frac { 1 } { \sec \theta } = ?

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Complete the identity. - csc⁡2θ2=?\csc ^ { 2 } \frac { \theta } { 2 } = ?

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Complete the identity. - tan⁡4θ−sec⁡4θ=\tan ^ { 4 } \theta - \sec ^ { 4 } \theta = ?

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Solve the equation on the interval 0≤θ<2π.0 \leq \theta < 2 \pi . - sin⁡2θ+sin⁡θ=0\sin ^ { 2 } \theta + \sin \theta = 0

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Establish the identity. - cos⁡(x−y)−cos⁡(x+y)=2sin⁡xsin⁡y\cos ( x - y ) - \cos ( x + y ) = 2 \sin x \sin y

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Complete the identity. - sin⁡(4θ)=?\sin ( 4 \theta ) = ?

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