Precalculus (10th Edition)

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

Chapter 6 - Trigonometric Functions - 6.3 Properties of the Trigonometric Functions - 6.3 Assess Your Understanding - Page 395: 86



Work Step by Step

I know that $\cos$ has a period of $2\pi$, hence $\cos{\theta}=\cos{(\theta-2\pi)}.$ Thus $\cos{\frac{37\pi}{18}}=\cos{(\frac{37\pi}{18}-2\pi)}=\cos{\frac{\pi}{18}}$. I know that $\sec$ is an even function, which means $f(-\theta)=f(\theta).$ Therefore $\sec{-\frac{\pi}{18}}=\sec{\frac{\pi}{18}}$. Thus: $\cos{\frac{37\pi}{18}}\sec{\frac{-\pi}{18}}=\cos{\frac{\pi}{18}}\sec{\frac{\pi}{18}}=1$. Because $\cos{\theta}\sec{\theta}=1$.
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