Answer
$-\frac{\pi}{6}$
Work Step by Step
$\sin^{-1}{-\frac{1}{2}}=-\frac{\pi}{6}$, because $\sin{-\frac{\pi}{6}}=-\frac{1}{2}$ and $-\frac{\pi}{6}$ is in the range of $\sin^{-1}{x}$, which is $[-\frac{\pi}{2},\frac{\pi}{2}].$
Precalculus (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0-32197-907-9
ISBN 13:
978-0-32197-907-0
Chapter 9 - Polar Coordinates; Vectors - Chapter Review - Cumulative Review - Page 637: 9
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