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: 19

Answer

$\dfrac{2\sqrt3}{2}$

Work Step by Step

Note that $\sin^{-1}{\left(-\dfrac{1}{2}\right)}=-\dfrac{\pi}{6}$, because $\sin{\left(-\dfrac{\pi}{6}\right)}=-\dfrac{1}{2}$ and $-\dfrac{\pi}{6}$ is in the range of $\sin^{-1}{x}$, which is $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right].$ Thus $\sec{\left(\sin^{-1}{\left(-\dfrac{1}{2}\right)}\right)}=\sec{\left(-\dfrac{\pi}{6}\right)}=\dfrac{2\sqrt3}{2}$
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