Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 6 - Analytic Trigonometry - Section 6.2 The Inverse Trigonometric Functions (Continued) - 6.2 Assess Your Understanding - Page 480: 9



Work Step by Step

The range of $\sin^{-1}{x}$ is $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$. Here, $\sin^{-1}{\left(\dfrac{\sqrt2}{2}\right)}=\dfrac{\pi}{4}$, because $\sin{\left(\dfrac{\pi}{4}\right)}=\dfrac{\sqrt2}{2}$ and $\dfrac{\pi}{4}$ lies in the range of $\sin^{-1}{x}$ . Therefore, we have: $\cos{\left(\sin^{-1}{\left(\dfrac{\sqrt2}{2}\right)}\right)}=\cos ({\dfrac{\pi}{4}})=\dfrac{\sqrt2}{2}$
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