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Let cos(−θ)=4/5 and tanθ> 0. What is the value of sin(−θ)?

User Sussie
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2 Answers

4 votes

Answer:


-(3)/(5)

Explanation:

Let cos(−θ)=4/5 and tanθ> 0. What is the value of sin(−θ)?-example-1
User Jason Mock
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5.1k points
4 votes
ANSWER

{ \sin ( \theta) } = -(3)/(5)

EXPLANATION

The cosine function is an even function.

It has the following property.


\cos( - \theta) = \cos( \theta)

This implied that if


\cos( - \theta) = (4)/(5)

Then,


\cos( \theta) = (4)/(5)

Using the Pythagorean identity,


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

We substitute the value of

\cos( \theta) = (4)/(5)

into the equation to get,


( { (4)/(5) })^(2) + { \sin ^(2) ( \theta) } = 1


(16)/(25)+ { \sin ^(2) ( \theta) } = 1


{ \sin ^(2) ( \theta) } = 1 - (16)/(25)


{ \sin ^(2) ( \theta) } = (9)/(25)

We take square root of both sides to get,


{ \sin ( \theta) } = \pm \: \sqrt{ (9)/(25) }


{ \sin ( \theta) } = \pm (3)/(5)

It was given that,


\tan( \theta) \: > \: 0

This implies that the angle is in the first quadrant. That is the only quadrant where both the cosine and the tangent ratios are positive.

Hence,


{ \sin ( \theta) } = (3)/(5).

But the sine function is an odd function.

This means that,


\sin( - \theta) =-\sin( \theta)

Therefore,


{ \sin ( -\theta) } =- (3)/(5).
User Ronny Bigler
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4.8k points