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: 16

Answer

$-\dfrac {\sqrt {33}}{6}$

Work Step by Step

$\cos \left( -\theta \right) =\cos \theta =\dfrac {\sqrt {3}}{6}$ (cosine is positive) $\cot \theta < 0$ ($\cot $ is negative) So angle is located in quadrant 4 and sine is negative in quadrant 4 $\Rightarrow \sin \theta =-\sqrt {1-\cos ^{2}\theta }=-\sqrt {1-\left( \dfrac {\sqrt {3}}{6}\right) ^{2}}=-\dfrac {\sqrt {33}}{6}$
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