Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 7 - Analytic Trigonometry - 7.2 The Inverse Trigonometric Functions (Continued) - 7.2 Assess Your Understanding - Page 458: 70



Work Step by Step

$f^{-1}(g(\frac{5\pi}{6}))=\sin^{-1}(\cos{\frac{5\pi}{6}})$. $\cos{\frac{5\pi}{6}}=-\frac{\sqrt3}{2}$. Thus $\sin^{-1}\left(\cos{\frac{5\pi}{6}}\right)=\sin^{-1}{-\frac{\sqrt3}{2}}=-\frac{\pi}{3}$, because $\sin{-\frac{\pi}{3}}=-\frac{\sqrt3}{2}$ and $-\frac{\pi}{3}$ is in the range of $\sin^{-1}{x}$, which is $[-\frac{\pi}{2},\frac{\pi}{2}].$
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