Exam 6: Analytic Trigonometry

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Complete the identity. - (sin⁡θ+cos⁡θ)21+2sin⁡θcos⁡θ= ? \frac { ( \sin \theta + \cos \theta ) ^ { 2 } } { 1 + 2 \sin \theta \cos \theta } = \text { ? }

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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. - cos⁡θ=55,0<θ<π2\cos \theta = \frac { \sqrt { 5 } } { 5 } , \quad 0 < \theta < \frac { \pi } { 2 } \quad Find sin⁡(2θ)\sin ( 2 \theta )

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Express the sum or difference as a product of sines and/or cosines. - cos⁡(4θ)−cos⁡(6θ)\cos ( 4 \theta ) - \cos ( 6 \theta )

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

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Find the exact value of the expression. - sin⁡π12\sin \frac { \pi } { 12 }

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Simplify the expression. - cos⁡24x−sin⁡24x\cos ^ { 2 } 4 x - \sin ^ { 2 } 4 x

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Solve the equation on the interval 0≤θ<2π0 \leq \theta < 2 \pi - tan⁡(2θ)−tan⁡θ=0\tan ( 2 \theta ) - \tan \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=14,tan⁡θ>0\sin \theta = \frac { 1 } { 4 } , \quad \tan \theta > 0 \quad Find cos⁡θ2\cos \frac { \theta } { 2 }

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Find the exact value of the expression. - sin⁡10∘cos⁡50∘+cos⁡10∘sin⁡50∘\sin 10 ^ { \circ } \cos 50 ^ { \circ } + \cos 10 ^ { \circ } \sin 50 ^ { \circ }

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Solve the equation on the interval 0≤θ<2π0 \leq \theta < 2 \pi - cot⁡(2θ−π2)=1\cot \left( 2 \theta - \frac { \pi } { 2 } \right) = 1

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Find the exact value of the expression. - tan⁡10∘+tan⁡20∘1−tan⁡10∘tan⁡20∘\frac { \tan 10 ^ { \circ } + \tan 20 ^ { \circ } } { 1 - \tan 10 ^ { \circ } \tan 20 ^ { \circ } }

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Use a calculator to find the value of the expression rounded to two decimal places. - sin⁡−1(−17)\sin ^ { - 1 } \left( - \frac { 1 } { 7 } \right)

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

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Solve the problem. -The altitude of a projectile in feet (neglecting air resistance) is given by y=(tan⁡θ)x−16v2cos⁡2θx2y = ( \tan \theta ) x - \frac { 16 } { v ^ { 2 } \cos ^ { 2 } \theta } x ^ { 2 } where x is the horizontal distance covered in feet and v is the initial velocity of the projectile at an angle θ from the horizontal. Find the firing angle (in degrees) of a projectile fired at an initial velocity of 100 feet per second so That it strikes the ground 312.5 feet from the firing point.

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Establish the identity. - cot⁡2xcsc⁡x−1=1+sin⁡xsin⁡x\frac { \cot ^ { 2 } x } { \csc x - 1 } = \frac { 1 + \sin x } { \sin x }

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Find the exact value of the expression. Do not use a calculator. - tan⁡−1[tan⁡(−π8)]\tan ^ { - 1 } \left[ \tan \left( - \frac { \pi } { 8 } \right) \right]

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Complete the identity. - ln⁡∣cot⁡x∣=?\ln | \cot x | = ?

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Complete the identity. - csc⁡(2θ)−sec⁡(2θ)(tan⁡θ−1)=\csc ( 2 \theta ) - \sec ( 2 \theta ) ( \tan \theta - 1 ) = ?

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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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Solve the problem. -Find cos⁡θ2\cos \frac { \theta } { 2 } , given that cos⁡θ=14\cos \theta = \frac { 1 } { 4 } and θ\theta terminates in 0<θ<π/20 < \theta < \pi / 2 .

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