Answer
$\dfrac{\pi}{2}$
Work Step by Step
We are given the expression: $\cos^{-1} (\sin 3\pi)$
The period of $\sin x$ is $2\pi$:
$\cos^{-1} (\sin 3\pi)=\cos^{-1} (\sin \pi)=\cos^{-1} 0$
$y=\cos^{-1} x$ is the value of $y$ so that $\cos y=x$, where $0\leq y\leq \pi$.
From the unit circle, we have:
$x=\cos\left(\dfrac{\pi}{2}\right)=0$
The value of $y$ for which $x=0$ in the interval $[0,\pi]$ is $\dfrac{\pi}{2}$. So:
$\cos^{-1}(\sin 3\pi)=\dfrac{\pi}{2}$
