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consider this equation sin(theta) = -4 square root of 29/29 if theta is an angle in quadrant IV what is the value of cos (theta)

User Minus One
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1 Answer

4 votes

Answer:


\cos(\theta)= (√(377))/(29)

Explanation:

Given


\sin(\theta) = -(4)/(√(29)) -- the correct expression

Required


\cos(\theta)

We know that:


\sin^2(\theta) + \cos^2(\theta)= 1

Make
\cos^2(\theta) the subject


\cos^2(\theta)= 1 - \sin^2(\theta)

Substitute:
\sin(\theta) = -(4)/(√(29))


\cos^2(\theta)= 1 - (-(4)/(√(29)))^2

Evaluate all squares


\cos^2(\theta)= 1 - ((16)/(29))

Take LCM


\cos^2(\theta)= (29 - 16)/(29)


\cos^2(\theta)= (13)/(29)

Take square roots of both sides


\cos(\theta)= \±\sqrt{(13)/(29)}

cosine is positive in the 4th quadrant;

So:


\cos(\theta)= \sqrt{(13)/(29)}

Split


\cos(\theta)= (√(13))/(√(29))

Rationalize


\cos(\theta)= (√(13))/(√(29)) * (√(29))/(√(29))


\cos(\theta)= (√(13*29))/(29)


\cos(\theta)= (√(377))/(29)

User NKorotkov
by
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