Precalculus (6th Edition)

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

Chapter 7 - Trigonometric Identities and Equations - 7.7 Equations Involving Inverse Trigonometric Functions - 7.7 Exercises: 2

Answer

A

Work Step by Step

A. From the definition of arcsin, x is the number from $[-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$ for which $\displaystyle \sin x=\frac{\sqrt{2}}{2}.$ $\displaystyle \frac{\pi}{4}\in [-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$ and $\displaystyle \sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}$ so $\displaystyle \frac{\pi}{4}$ is the solution. B. From the definition of arccos, x is the number from $[0, \pi]$ for which $\displaystyle \cos x=-\frac{\sqrt{2}}{2}.$ $\displaystyle \frac{\pi}{4}\in [0, \pi] $and $\displaystyle \cos\frac{\pi}{4}=+\frac{\sqrt{2}}{2}$ so $\displaystyle \frac{\pi}{4}$ is NOT the solution ($\displaystyle \frac{3\pi}{4}$ is). C. From the definition of arctan, x is the number from $(-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2})$ for which $\displaystyle \tan x=\frac{\sqrt{3}}{3}.$ $\displaystyle \frac{\pi}{4}\in (-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}) $and $\displaystyle \tan\frac{\pi}{4}=$1 so $\displaystyle \frac{\pi}{4}$ is NOT the solution ($\displaystyle \frac{\pi}{6}$ is)
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