Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - Chapter 7 Test Prep - Review Exercises: 102

Answer

$-\dfrac {9\sqrt {2}}{8}$

Work Step by Step

$sec\left( 2\sin ^{-1}\left( -\dfrac {1}{3}\right) \right) =\dfrac {1}{\sin \left( 2\sin ^{-1}\left( -\dfrac {1}{3}\right) \right) }=\frac{1}{2\sin \left( \sin ^{-1}\left( -\dfrac {1}{3}\right) \right) \cos \left( \sin ^{-1}\left( -\dfrac {1}{3}\right) \right)} =\dfrac {1}{2\times \left( -\dfrac {1}{3}\right) \times \sqrt {1-\left( -\dfrac {1}{3}\right) ^{2}}}=\dfrac {1}{-\dfrac {4\sqrt {2}}{9}}=-\dfrac {9\sqrt {2}}{8}$
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