Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.1 Inverse Circular Functions - 6.1 Exercises - Page 264: 31


The exact value of $y$ in this case is $$y=\frac{\pi}{6}$$

Work Step by Step

$\DeclareMathOperator{\as}{arcsec}$ $$y=\as \frac{2\sqrt 3}{3}$$ First, we see that the domain of inverse secant function is $(-\infty,\infty)$. Therefore, in fact when we deal with inverse secant function, we do not need to do this checking step. The range of inverse secant function is $[0,\frac{\pi}{2})\hspace{0.2cm}U\hspace{0.2cm}(\frac{\pi}{2},\pi]$. In other words, $y\in[0,\frac{\pi}{2})\hspace{0.2cm}U\hspace{0.2cm}(\frac{\pi}{2},\pi]$. We can rewrite $y=\as\frac{2\sqrt 3}{3}$ into $\sec y=\frac{2\sqrt 3}{3}$ We know that $$\sec\frac{\pi}{6}=\frac{1}{\cos\frac{\pi}{6}}=\frac{1}{\frac{\sqrt 3}{2}}=\frac{2}{\sqrt3}=\frac{2\sqrt 3}{3}$$ And $\frac{\pi}{6}$ belongs to the range $[0,\frac{\pi}{2})\hspace{0.2cm}U\hspace{0.2cm}(\frac{\pi}{2},\pi]$. Therefore, the exact value of $y$ here is $$y=\frac{\pi}{6}$$
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