Answer
$ -\frac{\pi}{6}$
Work Step by Step
$sin^{-1}(cos\frac{2\pi}{3})=sin^{-1}(-cos\frac{\pi}{3})=sin^{-1}(-sin(\frac{\pi}{2}-\frac{\pi}{3}))=sin^{-1}(-sin\frac{\pi}{6})=-\frac{\pi}{6}$.
Precalculus (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0-32197-907-9
ISBN 13:
978-0-32197-907-0
Chapter 7 - Analytic Trigonometry - Chapter Review - Review Exercises - Page 505: 18
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