Precalculus (6th Edition)

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

Chapter 7 - Trigonometric Identities and Equations - 7.1 Fundamental Identities - 7.1 Exercises - Page 658: 20


$-\dfrac {\sqrt {45}}{7}$

Work Step by Step

$sec\theta =\dfrac {1}{\cos \theta }=\dfrac {7}{2}$ (positive) $\tan \theta < 0$ (negative) So angle is located in quadrant 4 ( sine is negative in quadrant 4) $\Rightarrow \sin \theta =-\sqrt {1-\cos ^{2}\theta }=-\sqrt {1-\left( \dfrac {2}{7}\right) ^{2}}=-\dfrac {\sqrt {45}}{7}$
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