Exam 6: Analytic Trigonometry

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Establish the identity. - cot⁡2x=(csc⁡x−1)(csc⁡x+1)\cot ^ { 2 } x = ( \csc x - 1 ) ( \csc x + 1 )

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Find the domain of the function f and of its inverse function f−1\mathrm { f } ^ { - 1 } . - f(x)=−5cos⁡(9x)f ( x ) = - 5 \cos ( 9 x )

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Establish the identity. - csc⁡(π2+u)=sec⁡u\csc \left( \frac { \pi } { 2 } + u \right) = \sec u

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Establish the identity. - sec⁡4x−tan⁡4x=sec⁡2x+tan⁡2x\sec ^ { 4 } x - \tan 4 x = \sec ^ { 2 } x + \tan ^ { 2 } x

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Complete the identity. - sec⁡4θ+sec⁡2θtan⁡2θ−2tan⁡4θ=\sec ^ { 4 } \theta + \sec ^ { 2 } \theta \tan ^ { 2 } \theta - 2 \tan ^ { 4 } \theta = ?

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Solve the equation. Give a general formula for all the solutions. - tan⁡θ=−1\tan \theta = - 1

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Find the exact value of the expression. - sin⁡25∘cos⁡95∘+cos⁡25∘sin⁡95∘\sin 25 ^ { \circ } \cos 95 ^ { \circ } + \cos 25 ^ { \circ } \sin 95 ^ { \circ }

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Complete the identity. - sin⁡(π−θ)=?\sin ( \pi - \theta ) = ?

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Use a calculator to find the value of the expression in radian measure rounded to two decimal places. - csc⁡−1(−74)\csc ^ { - 1 } \left( - \frac { 7 } { 4 } \right)

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Find the exact value of the expression. - tan⁡175∘−tan⁡55∘1+tan⁡175∘tan⁡55∘\frac { \tan 175 ^ { \circ } - \tan 55 ^ { \circ } } { 1 + \tan 175 ^ { \circ } \tan 55 ^ { \circ } }

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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. - 2x2−3xsin⁡x=22 x ^ { 2 } - 3 x \sin x = 2

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Use the information given about the angle θ,0≤θ≤2π\theta , 0 \leq \theta \leq 2 \pi , to find the exact value of the indicated trigonometric function. - sin⁡θ=2425,0<θ<π2\sin \theta = \frac { 24 } { 25 } , \quad 0 < \theta < \frac { \pi } { 2 } \quad Find cos⁡(2θ)\cos ( 2 \theta ) .

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Find the exact value of the expression.884:890 - cos⁡[2sin⁡−1(−513)]\cos \left[ 2 \sin ^ { - 1 } \left( - \frac { 5 } { 13 } \right) \right]

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Solve the problem. -A mass hangs from a spring which oscillates up and down. The position P (in feet) of the mass at time t (in seconds) is given by P=4cos⁡(4t)P = 4 \cos ( 4 t ) For what values of t,0≤t<πt , 0 \leq t < \pi , will the position be 222 \sqrt { 2 } feet? Find the exact values. Do not use a calculator.

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Solve the problem. -Find sin⁡θ2\sin \frac { \theta } { 2 } , given that sin⁡θ=−35\sin \theta = - \frac { 3 } { 5 } and θ\theta terminates in 270∘<θ<360∘270 ^ { \circ } < \theta < 360 ^ { \circ } .

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Use a calculator to solve the equation on the interval 0 0≤θ<2π0 \leq \theta < 2 \pi . Round the answer to two decimal places. - sin⁡θ=−0.41\sin \theta = - 0.41

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Solve the equation. Give a general formula for all the solutions. - 2cos⁡θ+1=02 \cos \theta + 1 = 0

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Establish the identity. - cot⁡(x+y)cot⁡(x−y)=1−tan⁡2xtan⁡2ytan⁡2x−tan⁡2y\cot ( x + y ) \cot ( x - y ) = \frac { 1 - \tan ^ { 2 } x \tan ^ { 2 } y } { \tan ^ { 2 } x - \tan ^ { 2 } y }

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Express the sum or difference as a product of sines and/or cosines. - sin⁡(4θ)−sin⁡(2θ)\sin ( 4 \theta ) - \sin ( 2 \theta )

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Solve the equation on the interval 0≤θ<2π.0 \leq \theta < 2 \pi . - cos⁡2θ+2cos⁡θ+1=0\cos ^ { 2 } \theta + 2 \cos \theta + 1 = 0

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