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: 42



Work Step by Step

Solve for radians, then convert to degrees. $y=\sec^{-1}x =$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 $\sec y=-2$ $(\displaystyle \cos y=\frac{1}{-2}=-\frac{1}{2})$ In quadrant I, $\displaystyle \cos\frac{\pi}{3}=\frac{1}{2}$. In quadrant II, $\displaystyle \cos\frac{2\pi}{3}=-\frac{1}{2}$, that is, $\displaystyle \sec\frac{2\pi}{3}=-2, $ and $\displaystyle \frac{2\pi}{3}\in[0,\frac{\pi}{2})\cup(\frac{\pi}{2},\pi]$. So, $y=\displaystyle \frac{2\pi}{3}$ To convert radians to degrees, multiply y with $\displaystyle \frac{180^{o}}{\pi}$ $\displaystyle \theta=\frac{2\pi}{3}\cdot\frac{180^{o}}{\pi}=120^{o}$
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