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Solving Trigonometry functions

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1. Find the exact value of the sine, cosine, and tangent of the angle. 11pie/12

2. Find all solutions of the equation in the interval [0, 2pie).
Sin(x-5pie/6)-sin(x+ 5pie/6) =1

3. Find the exact value of sin2x using the double angle formula. Please help explain the use of the double angle formula.
Sin x=1/7, 0<x<pie/2

4. Given cos theta = 4/9, where 3pie/2 <or equal to theta <or equal to 2pie.

5. Express 2sin3xcos6x as a sum containing only sines or cosines.

6. Express cos7x-cos5x as a product containing only sines and/or cosines.

7. Evaluate the expressions without the aid of a calculator.

a. Arctan(- sqrt3/3)
b. Arcsin(-1/2)

8. Use inverse functions to evaluate the expressions,

a. cos(arcsin(sqrt5/5))
b. cos(arctan(sqrt2/x))

9. Identify the x-values that are solutions of the equations.

a. 8cos x-4 = 0
b. 18cot^2 x-18 = 0

10. A large pole is 175 feet tall. On a particular day at noon it casts a 198-foot shadow. What is the sun's angle of elevation?

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Solution Summary

I have provided step by step solutions to all the questions. Every trigonometry identity used has been typed out. This is a very good question and answer set on solving trig identities. Great practice questions as well.

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You need to refresh all your trig identities. There are plenty of web sites which talk about those. I have high lighted the trig identities that I used in each problem.

1. Find the exact value of the sine, cosine, and tangent of the angle. 11pie/12

Use : sin x = sin (p-x)

Sin (11p/12) = sin(p-11p/12) = sin(p/12)

Since we know what is sin pi/6 using the following identity makes sense.

Use: cos 2x = 1 - 2sin2 x è sinx = sqrt {½ (1-cos2x)}

Sin (11p/12) = sin(p-11p/12) = sin(p/12) = sqrt {½ (1-cos(p/6)}= Ö (1/2 (1-Ö3/2))
Sin (11p/12 = Ö(2-Ö3)/2 This is the exact value.

Evaluating we get Sin (11p/12) = 0.258819

2. Find all solutions of the equation in the interval [0, 2pie).
Sin(x-5pie/6)-sin(x+ 5pie/6) =1

Use the trig identity : sin C - sin D = 2 cos (C+D)/2 * sin (C-D)/2

Sin(x-5pie/6)-sin(x+ 5pie/6) = 2 cos x * sin -(5pi/6) = 1

2 cos x * sin -(5pi/6) = 1 è cos x * sin 5pi/6 = - 1/2,
we know sin 5pi/6 = sin pi - 5pi/6 = sin pi/6 = ½

cos x * 1/2 = -1/2, è cos x = -1

Between 0 and 2pi we have only one ...

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