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: 23



Work Step by Step

Note that $\cos{\left(-\dfrac{7\pi}{6}\right)}=-\dfrac{\sqrt3}{2}$. Thus $\sin^{-1}\left(\cos{\left(-\dfrac{7\pi}{6}\right)}\right)=\sin^{-1}{\left(-\dfrac{\sqrt3}{2}\right)}=-\dfrac{\pi}{3}$, because $\sin{\left(-\dfrac{\pi}{3}\right)}=-\dfrac{\sqrt3}{2}$ and $-\dfrac{\pi}{3}$ is in the range of $\sin^{-1}{x}$, which is $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right].$
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