Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $ It is known that
\begin{align}
\arcsin x + \arcsin y  =\begin{cases}
\arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \\\quad\text{if } x^2+y^2 \le 1 &\text{or} &(x^2+y^2 > 1 &\text{and} &xy< 0);\\
\pi - \arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \\\quad\text{if } x^2+y^2 > 1&\text{and} &0< x,y \le 1;\\
-\pi - \arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \\\quad\text{if } x^2+y^2 > 1&\text{and} &-1\le x,y < 0.
\end{cases}
\end{align}
I tried to prove this myself, have no problem in getting the 'crux' $\arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2})$ part of the RHS, but face trouble in checking the range of that 'crux' under the given conditions.

Take the sine of both sides, and use the angle addition formula, then further simplify it by using the fact that $\cos\arcsin t=\sqrt{\cos^2\arcsin t}=\sqrt{1-\sin^2\arcsin t}=\sqrt{1-t^2}$. Then apply the $\arcsin$ function to both sides, and you're done.
