Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.1 Inverse Circular Functions - 6.1 Exercises - Page 260: 82

Answer

$sec~(sin^{-1}~(-\frac{1}{5})) = \frac{5~\sqrt{24}}{24}$

Work Step by Step

$\theta = sin^{-1}(-\frac{1}{5})$ $sin~\theta = -\frac{1}{5} = \frac{opposite}{hypotenuse}$ Note that $\theta$ is in quadrant IV. We can find the magnitude of the adjacent side: $adjacent = \sqrt{5^2-1^2} = \sqrt{24}$ In quadrant IV, $sec~\theta$ is positive. We can find the value of $sec~\theta$: $sec~\theta = \frac{hypotenuse}{adjacent}$ $sec~\theta = \frac{5}{\sqrt{24}}$ $sec~\theta = \frac{5~\sqrt{24}}{24}$ Therefore, $sec~(sin^{-1}~(-\frac{1}{5})) = \frac{5~\sqrt{24}}{24}$
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