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: 32


$y=\displaystyle \frac{3\pi}{4}$

Work Step by Step

$y=\sec^{-1}x =$arcsec$x$ Domain: $(-\infty, -1]\cup[1, \infty)$ Range: $[0,\displaystyle \frac{\pi}{2})\cup(\frac{\pi}{2},\pi]$ Quadrants (unit circle): I and II -------- $y$ is the number from $[0,\displaystyle \frac{\pi}{2})\cup(\frac{\pi}{2},\pi]$ such that $\sec y=-\sqrt{2}$ $(\displaystyle \cos y=\frac{1}{-\sqrt{2}}=-\frac{\sqrt{2}}{2})$ In quadrant I, $\displaystyle \cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}$, and in quadrant II, $\displaystyle \cos\frac{3\pi}{4}=-\frac{\sqrt{2}}{2}$, that is, $\displaystyle \sec\frac{3\pi}{4}=-\sqrt{2}, $ and $\displaystyle \frac{3\pi}{4}\in[0,\frac{\pi}{2})\cup(\frac{\pi}{2},\pi]$. So, $y=\displaystyle \frac{3\pi}{4}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.