Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - 7.5 Inverse Circular Functions - 7.5 Exercises - Page 708: 14

Answer

$-\displaystyle \frac{\pi}{2}$

Work Step by Step

$y=\sin^{-1}x$ Domain: $[-1, 1] $Range: $[-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$ $y$ is the number from $[-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$ such that $\sin y=-1$ ---------------- $\displaystyle \sin(-\frac{\pi}{2})=-1$, and $-\displaystyle \frac{\pi}{2}\in[-\frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$ so $\displaystyle \sin^{-1}(-1)=-\frac{\pi}{2}$
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