Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 7 - Analytic Trigonometry - 7.2 The Inverse Trigonometric Functions (Continued) - 7.2 Assess Your Understanding - Page 458: 13



Work Step by Step

Note that $\cos^{-1}{\left(\dfrac{1}{2}\right)}=\dfrac{\pi}{3}$, because $\cos{\left(\dfrac{\pi}{3}\right)}=\dfrac{1}{2}$ and $\dfrac{\pi}{3}$ is in the range of $\cos^{-1}{x}$, which is $\left[0,\pi\right].$ Thus $\sec{\left(\cos^{-1}{\left(\dfrac{1}{2}\right)}\right)}=\sec{\left(\dfrac{\pi}{3}\right)}=2.$
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