<s>[INST] 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.
 [/INST] Let $a=\sin^{-1}x,$ $b=\sin^{-1}y\implies\sin a=x,$ $\sin b=y$ and $a,b\in[-\pi/2,\pi/2]\implies a+b\in[-\pi,\pi]$
$$
\sin(a+b)=\sin a\cos b+\cos a\sin b=x\sqrt{1-y^2}+y\sqrt{1-x^2}\\=\sin\bigg[\sin^{-1}\Big(x\sqrt{1-y^2}+y\sqrt{1-x^2}\Big)\bigg]\\
\implies a+b=\color{red}{\sin^{-1}x+\sin^{-1}y=n\pi+(-1)^n\sin^{-1}\Big(x\sqrt{1-y^2}+y\sqrt{1-x^2}\Big)}
$$
Case 1: $-\dfrac{\pi}{2}\leq\sin^{-1}x+\sin^{-1}y\leq\dfrac{\pi}{2}$
$$
\color{darkblue}{\sin^{-1}x+\sin^{-1}y=\sin^{-1}\Big(x\sqrt{1-y^2}+y\sqrt{1-x^2}\Big)}
$$
$$
\cos(a+b)\geq0\implies \cos a\cos b-\sin a\sin b\geq0\implies \cos a\cos b\geq\sin a\sin b\\
\sqrt{1-x^2}\sqrt{1-y^2}\geq xy\implies\\
a)\ 1-x^2-y^2+x^2y^2-x^2y^2\geq0 \implies x^2+y^2\leq1\\
b)\ 1-x^2-y^2+x^2y^2-x^2y^2<0 \implies x^2+y^2>1, xy<0
$$
Case 2 & 3: $\dfrac{\pi}{2}<\sin^{-1}x+\sin^{-1}y\leq\pi$ and $-\pi\leq\sin^{-1}x+\sin^{-1}y<-\dfrac{\pi}{2}$
$$
\cos(a+b)<0\implies x^2+y^2>1
$$
Case 2-: $\dfrac{\pi}{2}<\sin^{-1}x+\sin^{-1}y\leq\pi$
$$
\sin^{-1}\Big(x\sqrt{1-y^2}+y\sqrt{1-x^2}\Big)=\pi-(\sin^{-1}x+\sin^{-1}y)\\
\implies \color{darkblue}{\sin^{-1}x+\sin^{-1}y=\pi-\sin^{-1}\Big(x\sqrt{1-y^2}+y\sqrt{1-x^2}\Big)}
$$
$$
\dfrac{\pi}{2}<\sin^{-1}x+\sin^{-1}y\leq\pi \;\&\;-\dfrac{\pi}{2}\leq\sin^{-1}x,\;\sin^{-1}y\leq\dfrac{\pi}{2}\\
\implies 0<\sin^{-1}x,\sin^{-1}y\leq\dfrac{\pi}{2}\implies 0< x,y\leq1
$$
Case 3-: $-\pi\leq\sin^{-1}x+\sin^{-1}y<-\dfrac{\pi}{2}$
$$
\sin^{-1}\Big(x\sqrt{1-y^2}+y\sqrt{1-x^2}\Big)=-\pi-(\sin^{-1}x+\sin^{-1}y)\\
\implies \color{darkblue}{\sin^{-1}x+\sin^{-1}y=-\pi-\sin^{-1}\Big(x\sqrt{1-y^2}+y\sqrt{1-x^2}\Big)}
$$
$$
-\pi\leq\sin^{-1}x+\sin^{-1}y<-\dfrac{\pi}{2} \;\&\;-\dfrac{\pi}{2}\leq\sin^{-1}x,\;\sin^{-1}y\leq\dfrac{\pi}{2}\\
\implies -\dfrac{\pi}{2}\leq\sin^{-1}x,\sin^{-1}y< 0\implies -1\leq x,y< 0
$$
</s>