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telo118 [61]
2 years ago
6

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

Mathematics
1 answer:
Deffense [45]2 years ago
6 0

Answer:    -\frac{240}{289}

=========================================================

Explanation:

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

Isolating sine leads to \sin(\theta) = -\frac{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(-\frac{15}{17}\right)*\left(\frac{8}{17}\right)\\\\\sin(2\theta) = -\frac{240}{289}\\\\

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