Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 3 - Section 3.1 - Reference Angle - 3.1 Problem Set - Page 122: 24


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

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 III, The reference angle = $ 180^{\circ} - 150^{\circ} $ = $30^{\circ}$ As $ (-150)^{\circ}$ terminates in quadrant III, its $\cos$ will be negative. Therefore by reference angle theorem- $\cos (-150)^{\circ}$ = - $\cos 30^{\circ}$ = - $\frac{\sqrt 3}{2}$
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