MHB Confirming $\arcsin(-\sin(\frac{3\pi}{2}))$

[Guest2]

What's $\arcsin(\sin(\frac{−3\pi}{2}))?$ I think it's equal to

$\arcsin(-\sin(\frac{3\pi}{2})) =\arcsin(-\sin(\pi+\frac{\pi}{2})) = \arcsin(\sin(\pi/2)) = \frac{\pi}{2}$

Is this correct?

 

[Prove It]

What's $\arcsin(\sin(\frac{−3\pi}{2}))?$ I think it's equal to

$\arcsin(-\sin(\frac{3\pi}{2})) =\arcsin(-\sin(\pi+\frac{\pi}{2})) = \arcsin(\sin(\pi/2)) = \frac{\pi}{2}$

Is this correct?

Yes it's correct, but if you realize that an angle of $\displaystyle \begin{align*} -\frac{3\,\pi}{2} \end{align*}$ radians is the same as an angle of $\displaystyle \begin{align*} \frac{\pi}{2} \end{align*}$ radians it saves about two lines of working...