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: 85



Work Step by Step

I know that $\csc$ has a period of $2\pi$, hence $\csc{\theta}=\tan{(\theta-2\pi)}.$ Thus $\csc{\frac{25\pi}{12}}=\csc{(\frac{25\pi}{12}-2\pi)}=\csc{\frac{\pi}{12}}$ I know that $\sin$ is an odd function, which means $f(-\theta)=-f(\theta).$ Therefore $\sin{-\frac{\pi}{12}}=-\sin{\frac{\pi}{12}}$. Thus: $\csc{\frac{25\pi}{12}}\sin{-\frac{\pi}{12}}=\csc{\frac{\pi}{12}}-\sin{\frac{\pi}{12}}=-1$. Because $\csc{\theta}\sin{\theta}=1$.
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