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: 48



Work Step by Step

Let $\theta = \sec^{-1} {-3} \hspace{20pt} \therefore \sec{\theta} = -3$ Since $\cos{\theta}= \dfrac{1}{\sec{\theta}}$, then $\cos{\theta} = -\dfrac{1}{3}$ Thus, $\theta = \cos^{-1} {\left(-\frac{1}{3}\right)}\approx 1.91$ Hence, $\sec^{-1} {-3} = \cos^{-1} {\left(-\frac{1}{3}\right)} \approx \boxed{1.91}$
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