# How do you evaluate arccos(sin((3*pi)/2))?

$\pi$
initially $\sin \left(3 \frac{\pi}{2}\right) = \sin \left(\pi + \frac{\pi}{2}\right) = - \sin \frac{\pi}{2} = - 1$
then $\arccos \left(\sin 3 \frac{\pi}{2}\right) = \arccos \left(- 1\right) = \arccos \left(\cos \pi\right) = \pi$