Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 6 - Analytic Trigonometry - Section 6.2 The Inverse Trigonometric Functions (Continued) - 6.2 Assess Your Understanding - Page 480: 41



Work Step by Step

We know that the trigonometric function $\sec^{-1} x$ has range: $[0, \pi]$. Here, $\sec^{-1} (\dfrac{2 \sqrt 3}{3})=\dfrac{\pi}{6}$, because $\sec(\dfrac{\pi}{6})=\dfrac{2 \sqrt 3}{3}$ and $\dfrac{\pi}{6}$ lies in the range of $\sec^{-1}{x}$ .
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