## Precalculus (6th Edition)

Published by Pearson

# Chapter 7 - Trigonometric Identities and Equations - 7.5 Inverse Circular Functions - 7.5 Exercises - Page 708: 40

#### Answer

$-45^{o}$

#### Work Step by Step

Solve for radians, then convert to degrees. $y=\sin^{-1}x =\arcsin x$ 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 $\displaystyle \sin y=-\frac{\sqrt{2}}{2}.$ In quadrant I, $\displaystyle \sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}.$ In quadrant IV, $\displaystyle \sin(-\frac{\pi}{4})=-\frac{\sqrt{2}}{2}\qquad$and$\displaystyle \quad -\frac{\pi}{4}\in[-\frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$, so $y =-\displaystyle \frac{\pi}{4}$ To convert radians to degrees, multiply y with $\displaystyle \frac{180^{o}}{\pi}$ $\displaystyle \theta=-\frac{\pi}{4}\cdot\frac{180^{o}}{\pi}=-45^{o}$

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