Exam 6: Analytic Trigonometry

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Find the exact value of the expression. - sin(cos112sin132)\sin \left( \cos ^ { - 1 } \frac { 1 } { 2 } - \sin ^ { - 1 } \frac { \sqrt { 3 } } { 2 } \right)

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Solve the equation on the interval 0θ<2π0 \leq \theta < 2 \pi - tanθ+secθ=1\tan \theta + \sec \theta = 1

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Solve the equation on the interval 0θ<2π0 \leq \theta < 2 \pi \text {. } - cosθ=sinθ\cos \theta = \sin \theta

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Solve the equation on the interval 0θ<2π0 \leq \theta < 2 \pi - 2cosθ+1=02 \cos \theta + 1 = 0

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Complete the identity. - sin(3θ)sinθ=?\frac { \sin ( 3 \theta ) } { \sin \theta } = ?

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Solve the equation. Give a general formula for all the solutions. - cscθ3=233\csc \frac { \theta } { 3 } = \frac { 2 \sqrt { 3 } } { 3 }

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Establish the identity. - tan(π2+x)=cotx\tan \left( \frac { \pi } { 2 } + x \right) = - \cot x

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Solve the problem. -A consumer notes the sinusoidal nature of her monthly power bills. In winter when she uses electricity to heat her home and in summer when she cools her home, the bills are high. In spring and fall, significantly less electricity is used and the bills are much smaller. The following function models this behavior. C=60+40cos(π3tπ3)C = 60 + 40 \cos \left( \frac { \pi } { 3 } t - \frac { \pi } { 3 } \right) Here C is the cost of power in dollars for the month t, 1 1t121 \leq t \leq 12 2, with t = 1 corresponding to January. For what values of t, 1 1t121 \leq t \leq 12 12, is the cost exactly $80?

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Solve the problem. -The path of a projectile fired at an inclination θ\theta (in degrees) to the horizontal with an initial speed v0 is a parabola. The range R of the projectile, that is, the horizontal distance that the projectile travels, is found by using the formula R=v02 gsin(2θ)R = \frac { \mathrm { v } _ { 0 } ^ { 2 } } { \mathrm {~g} } \sin ( 2 \theta ) where g is the acceleration due to gravity. The maximum height H of the projectile is H=v024 g(1cos(2θ))\mathrm { H } = \frac { \mathrm { v } _ { 0 } ^ { 2 } } { 4 \mathrm {~g} } ( 1 - \cos ( 2 \theta ) ) Find the range R and the maximum height H in terms of g if the projectile is fired with an initial speed of 200 meters per second at an angle of 15° and then at an angle of 22.5°. Do not use a calculator, but simplify the answers.

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Use the Half-angle Formulas to find the exact value of the trigonometric function. - cos5π12\cos \frac { 5 \pi } { 12 }

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Solve the equation. Give a general formula for all the solutions. - sinθ=1\sin \theta = 1

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Complete the identity. - cosθcosθsin2θ=\cos \theta - \cos \theta \sin ^ { 2 } \theta = ?

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Find the exact solution of the equation. - sin1x=π6\sin ^ { - 1 } x = \frac { \pi } { 6 }

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Complete the identity. - sin(2θ)+2sin2θcos(2θ)=?\frac { \sin ( 2 \theta ) + 2 \sin ^ { 2 } \theta } { \cos ( 2 \theta ) } = ?

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Use a calculator to solve the equation on the interval 0 0θ<2π0 \leq \theta < 2 \pi . Round the answer to two decimal places. - cosθ=0.92\cos \theta = 0.92

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Find the exact value of the expression. - sin15\sin 15 ^ { \circ }

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

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Use the information given about the angle θ,0θ2π\theta , 0 \leq \theta \leq 2 \pi , to find the exact value of the indicated trigonometric function. - cosθ=57,cscθ<0\cos \theta = - \frac { 5 } { 7 } , \quad \csc \theta < 0 \quad Find cos(2θ)\cos ( 2 \theta )

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Solve the problem. -Wildlife management personnel use predator-prey equations to model the populations of certain predators and their prey in the wild. Suppose the population M of a predator after t months is given by M=750+125sinπ6tM = 750 + 125 \sin \frac { \pi } { 6 } t while the population N of its primary prey is given by N=12,250+3050cosπ6tN = 12,250 + 3050 \cos \frac { \pi } { 6 } t Find the values of t 0t<120 \leq t < 12 or which the predator population is 875. Find the values of t, 0t<120 \leq t < 12 for which the prey population is 10,725.

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Find the exact value of the expression. Do not use a calculator. - sin1[sin(6π7)]\sin ^ { - 1 } \left[ \sin \left( \frac { 6 \pi } { 7 } \right) \right]

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