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: $-\frac{240}{289}$

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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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