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: 15


$=-\dfrac {2\sqrt {5}}{5}$

Work Step by Step

Cosine is even function so $\cos \left( -\theta \right) =\dfrac {\sqrt {5}}{5}\Rightarrow \cos \theta =\dfrac {\sqrt {5}}{5}$ We see that $\cos \theta $ is positive and $\tan \theta $ is negative so angle is located in quadrant 4 and sine function is negative in quadrant 4 $\Rightarrow \sin \theta =-\sqrt {1-\cos ^{2}\theta }=-\sqrt {1-\left( \dfrac {\sqrt {5}}{5}\right) ^{2}}=-\dfrac {2\sqrt {5}}{5}$
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