Answer
$-\dfrac{1}{2}$
Work Step by Step
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, $-\sin 30^{\circ}=-\dfrac{1}{2}$
