## Trigonometry 7th Edition

Published by Cengage Learning

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

#### Answer

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

#### 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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