Precalculus (10th Edition)

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

Chapter 7 - Trigonometry and Periodic Functions - 7.1 The Inverse Sine, Cosine, and Tangent Functions - 7.1 Assess Your Understanding - Page 451: 21



Work Step by Step

The domain of $\sin^{-1}{x}$ is the range of $\sin{x}$, which is $[−1, 1] $. The range of $\sin^{-1}{x}$ is a specified domain of $\sin{x}$, i.e. $\left[− \frac{\pi}{2} , \frac{\pi}{2} \right]$. We know that if $\sin y=x$, then $\sin^{-1}(x)=y.$ We know that $\sin(\frac{\pi}{4})=\frac{\sqrt2}{2}$, hence $\sin^{-1}(\frac{\sqrt2}{2})=\frac{\pi}{4}.$
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