Precalculus (6th Edition)

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

Chapter 5 - Trigonometric Functions - 5.3 Trigonometric Functions Values and Angle Measures - 5.3 Exercises - Page 533: 91

Answer

$-\frac{\sqrt 3}{2}$

Work Step by Step

As the period of cosine function is $360^{\circ}$, $\cos(-510^{\circ})=\cos(-510^{\circ}+360^{\circ})$ $=\cos(-150^{\circ})=\cos 150^{\circ}$ (As $\cos (-x)=\cos x$ ) $=\cos(180^{\circ}-30^{\circ})=-\cos 30^{\circ}$ ( As $\cos(180^{\circ}-x)=-\cos x$ ) $=-\frac{\sqrt 3}{2}$
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