Algebra 2 (1st Edition)

$-\dfrac{1}{2}$
When $P$ is the period of $f(x)$ then we have $f(x+p)=f(x)$ Here, we have $\sin \theta=-150^{\circ}$ and $\sin (-150^{\circ})=\sin [(-150^{\circ})^{\circ}+360^{\circ})=\sin 210^{\circ}$ and $\theta'= 210^{\circ}- 180^{\circ}= 30^{\circ}$ $\sin 30^{\circ}=\dfrac{1}{2}$ Since the function sine is negative in the third quadrant: Thus, $-\sin 30^{\circ}=-\dfrac{1}{2}$