## Precalculus (10th Edition)

a)$\frac{1}{3}$ b) $1$
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 $f(a)+f(a+2\pi)+f(a+4\pi)=\sin{(a)}+\sin{(a+2\pi)}+\sin{(a+4\pi)}=\sin{(a)}+\sin{(a)}+\sin{(a)}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1$