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: 26



Work Step by Step

We know that $sec$ has a period of $2\pi$, hence first we try and find a value where the argument is between $\frac{-\pi}{2}$ and $\frac{3\pi}{2}$. Therefore $sec(8\pi)=sec(8\pi-2\pi)=sec(6\pi)=sec(6\pi-2\pi)=sec(4\pi)=sec(4\pi-2\pi)=sec(2\pi)=sec(2\pi-2\pi)=sec(0)=1$
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