Precalculus (6th Edition)

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

Chapter 7 - Trigonometric Identities and Equations - 7.5 Inverse Circular Functions - 7.5 Exercises - Page 709: 78

Answer

$$\frac{5}{{2\sqrt 6 }}$$

Work Step by Step

$$\eqalign{ & \sec \left( {{{\sin }^{ - 1}}\left( { - \frac{1}{5}} \right)} \right) \cr & = \sec \left( {{{\sin }^{ - 1}}\left( {\frac{1}{5}} \right)} \right) \cr & {\text{Let }}\theta = {\sin ^{ - 1}}\left( {\frac{1}{5}} \right){\text{, thus}} \cr & \sin \theta = \frac{1}{5} \cr & {\text{Recall that }}\sin \theta = \frac{{{\text{opposite side}}}}{{{\text{hypotenuse}}}} \cr & {\text{opposite side}} = 1 \cr & {\text{hypotenuse}} = 5 \cr & {\text{adjacent side}} = \sqrt {25 - 1} = 2\sqrt 6 \cr & {\text{Then}} \cr & \sec \theta = \sec \left( {{{\sin }^{ - 1}}\left( {\frac{1}{5}} \right)} \right) = \frac{{{\text{hypotenuse}}}}{{{\text{adjacent side}}}} \cr & \sec \left( {{{\sin }^{ - 1}}\left( { - \frac{1}{5}} \right)} \right) = \frac{5}{{2\sqrt 6 }} \cr} $$
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