Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 6 - Trigonometric Functions - 6.2 Trigonometric Functions: Unit Circle Approach - 6.2 Assess Your Understanding - Page 379: 22



Work Step by Step

We know that $cos$ has a period of $2\pi$, hence first we try and find a value where the argument is between $-\pi$ and $/pi$. Therefore $cos(7\pi)=cos (7\pi-2\pi)=cos (5\pi)=cos (5\pi-2\pi)=cos (3\pi)=cos (3\pi-2\pi)=cos (\pi)=-1$
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