Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 6 - Section 6.3 - Trigonometric Functions of Angles - 6.3 Exercises - Page 499: 13


Exact value of $\cos 150^{\circ}$ = - $\frac{\sqrt 3}{2}$

Work Step by Step

To find exact value of $\cos 150^{\circ}$, let's find its reference angle first. As $ 150^{\circ}$ terminates in quadrant II, The reference angle = $ 180^{\circ} - 150^{\circ}$ = $30^{\circ}$ As $ 150^{\circ}$ terminates in quadrant II, its cos will be negative. Therefore by reference angle theorem- $\cos 150^{\circ}$ = - $\cos30^{\circ}$ = - $\frac{\sqrt 3}{2}$
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