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: 113


a)$\frac{1}{3}$ b) $1$

Work Step by Step

a) I know that $\sin$ is an odd function, which means $f(-\theta)=-f(\theta).$ Therefore $f(-a)=\sin{(-a)}=-\sin{a}=-\frac{1}{3}$. b) I know that $\sin$ has a period of $2\pi$, hence $\sin{\theta}=\sin{(\theta+2k\pi)}$ where $k$ is an integer. Thus $f(a)+f(a+2\pi)+f(a+4\pi)=\sin{(a)}+\sin{(a+2\pi)}+\sin{(a+4\pi)}=\sin{(a)}+\sin{(a)}+\sin{(a)}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1$
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