So I was trying to solve this problem arcsin(sin(-3pi/5)

and so far I got that the arcsin(sin) can't cancel because the -3pi/5 is out the range of pi/2<x<-pi/2

So I looked for the reference angle...
and I know that -3pi/5 will lay on QIII because is going clockwise.
So I looked at the theta hat, and I see that theta hat is -2pi/5..

Is this right... or do I have a mistake... please help. I kind it feel that I did something wrong.

I would use the following:

$\sin(-x)=-\sin(x)$

$\sin^{-1}(-x)=-\sin^{-1}(x)$

$\sin(\pi-x)=\sin(x)$

to obtain:

$\sin^{-1}\left(\sin\left(-\frac{3\pi}{5} \right) \right)=-\sin^{-1}\left(\sin\left(\frac{2\pi}{5} \right) \right)=-\frac{2\pi}{5}$