Question #872b5

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Question #872b5

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Answer 1 · 2017-10-01T22:52:45

If $cos a = - \frac{1}{3}$ $cosa=-1/3$ , $a$ $a$ must be between $90$ $90$ and $180$ $180$ degrees. In other words, a in $2$ $2$ nd quadrant in unit circle. In this quadrant $sin a > 0$ $sina>0$ . So, if $cos a = - \frac{1}{3}$ $cosa=-1/3$ , $sin a = \frac{2 \sqrt{2}}{3}$ $sina=(2sqrt(2))/3$

After using $sin 2 a = 2 sin a \cdot cos a$ $sin2a=2sina*cosa$ identity, I found $sin 2 a = - \frac{4 \sqrt{2}}{9}$ $sin2a=-(4sqrt(2))/9$

Explanation:

1) I found $sin a$ $sina$ after being given $cos u = - \frac{1}{3}$ $cosu=-1/3$ and $90 < a < 180$ $90 < a < 180$ conditions.

2) I used $sin 2 a = 2 sin a \cdot cos a$ $sin2a=2sina*cosa$ identity.

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Question #872b5

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