Precalculus (6th Edition)

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

Chapter 6 - The Circular Functions and Their Graphs - Chapter 6 Test Prep - Review Exercises - Page 644: 32

Answer

$$ - \frac{1}{2}$$

Work Step by Step

$$\eqalign{ & \cos \frac{{2\pi }}{3} \cr & {\text{Write }}\frac{{2\pi }}{3}{\text{ as }}\pi - \frac{\pi }{3} \cr & \cos \frac{{2\pi }}{3} = \cos \left( {\pi - \frac{\pi }{3}} \right) \cr & {\text{Use the identity cos}}\left( {\pi - \theta } \right) = - \cos \theta \cr & \cos \frac{{2\pi }}{3} = - \cos \left( {\frac{\pi }{3}} \right) \cr & {\text{From the unit circle we known that cos}}\frac{\pi }{3} = \frac{1}{2},{\text{ then}} \cr & \cos \frac{{2\pi }}{3} = - \frac{1}{2} \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.