Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Review Exercises - Page 291: 32


$sec~(2~sin^{-1}~(-\frac{1}{3})) = \frac{9}{7}$

Work Step by Step

Let $\theta = sin^{-1}~(-\frac{1}{3})$ Then $sin~\theta = -\frac{1}{3}$ We can find the value of $sec~2\theta$: $sec~2\theta = \frac{1}{cos~2\theta}$ $sec~2\theta = \frac{1}{1-2sin^2~\theta}$ $sec~2\theta = \frac{1}{1-2(-\frac{1}{3})^2}$ $sec~2\theta = \frac{1}{1-(\frac{2}{9})}$ $sec~2\theta = \frac{1}{(\frac{7}{9})}$ $sec~2\theta = \frac{9}{7}$ $sec~(2~sin^{-1}~(-\frac{1}{3})) = \frac{9}{7}$
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