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26 votes
26 votes
Given that cos(theta) = 8/17 and that theta lies in Quadrant IV, what is the exact value of sin 2(theta)?

User Peterept
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1 Answer

17 votes
17 votes

Answer:
-(240)/(289)

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Step-by-step explanation:

Use the pythagorean trig identity
\sin^2(\theta)+\cos^2(\theta) = 1 and plug in the fact that
\cos(\theta) = (8)/(17)\\\\

Isolating sine leads to
\sin(\theta) = -(15)/(17)\\\\. I'm skipping the steps here, but let me know if you need to see them.

The result is negative because we're in quadrant 4, when y < 0 so it's when sine is negative.

Therefore,


\sin(2\theta) = 2\sin(\theta)\cos(\theta)\\\\\sin(2\theta) = 2*\left(-(15)/(17)\right)*\left((8)/(17)\right)\\\\\sin(2\theta) = -(240)/(289)\\\\

User Santhosh
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3.2k points