A geometric point of view might be illuminating.
Suppose $0\leq x \leq \frac{1}{\sqrt 2}$. Consider the figure below, where we start from a right-angled triangle $\triangle ABC$ with sides $\overline{AB} = \sqrt{1-x^2}$ and $\overline{BC} = x$, and hypotenuse $\overline{AC}=1$. The choice of $x$ we have made guarantees that
$$ \alpha = \angle BAC$$
is in the range $\left[0, \frac{\pi}{4}\right].$
By definition is
$$ \alpha = \arcsin x.$$
Extend first $CB$ to a segment $BD \cong BC$, then draw from $D$ the line perpendicular to $AD$ that intersects the extension of $AC$ in $E$. Finally draw from $C$ the perpendicular to $BC$ that meets $ED$ in $F$.
Define
\begin{equation}\beta = \angle CAD = 2\alpha.\tag{1}\label{eq:1}\end{equation}
We have, by definition,
\begin{equation}\beta = \arcsin \left(\frac{\overline{ED}}{\overline{AE}}\right).\tag{2}\label{eq:2}\end{equation}
From $\triangle DCF \sim \triangle ABC$ find
$$\overline{CF} = \frac{2x^2}{\sqrt{1-x^2}}$$
and
$$\overline{DF} = \frac{2x}{\sqrt{1-x^2}}.$$
By Pythagorean Theorem on $\triangle ADE$, and by $\triangle CEF \sim \triangle CED$
\begin{equation}
\begin{cases}
1+\overline{ED}^2 = (1+ \overline{EC})^2\\
\overline{EC} = \frac{x}{\sqrt{1-x^2}}\overline{ED}.
\end{cases}
\end{equation}
Solving the system yields
$$\overline{ED} = \frac{2x\sqrt{1-x^2}}{1-2x^2}$$
and
$$\overline{AC} = 1 + \overline{EC} = \frac{1}{1-2x^2}.$$
Using these results in \eqref{eq:2} and then cosidering the identity \eqref{eq:1} leads to
$$2\arcsin x = \arcsin \left(2x\sqrt{1-x^2}\right).$$
For $\frac{1}{\sqrt 2} \leq x \leq 1$, I would consider the triangle below, where again $\overline{BC} = x$, $\overline{AC} = 1$ and $D$ is the point symmetrical to $C$ with respect to line $AB$. Draw then from $C$ the perpendicular to $AD$ that meets its extension in $E$. Define then $\alpha$ as before and
$$\beta = \angle CAE = \pi - \angle CAD = \pi -2\alpha.$$
Use then the fact that $\sin \beta = \sin 2\alpha$.
Finally, for negative $x$ just define $\overline{BC} = -x$ and proceed as above, taking advantage of the sine odd symmetry.
Let's call $V$ the point where the violet beam exits the plastic, and $R$ where the red beam exits the plastic. Draw a perpendicular from $R$ to the outgoing violet beam, and call $P$ the point of intersection. In this right angle triangle, the hypotenuse is $h_{red}-h_{violet}$ and one of the sides is $D$. Notice that all outgoing beams will exit the plastic at the same angle with respect to the normal, so at $50^\circ$ with respect to the normal to the surface. Then the angle between the beam and the direction parallel to the surface is $40^\circ$. So in the $\triangle VRP$ you can write $$\sin 40^\circ=\frac{D}{h_{red}-h_{violet}}$$
Note that $\sin 40^\circ=\cos 50^\circ$, so $$D=(h_{red}-h_{violet})\cos50^\circ$$
Best Answer
Using this, $\displaystyle-\frac\pi2\leq \arcsin z\le\frac\pi2 $ for $-1\le z\le1$
So, $\displaystyle-\pi\le\arcsin x+\arcsin y\le\pi$
Again, $\displaystyle\arcsin x+\arcsin y= \begin{cases} \\-\pi- \arcsin(x\sqrt{1-y^2}+y\sqrt{1-x^2})& \mbox{if } -\pi\le\arcsin x+\arcsin y<-\frac\pi2\\ \arcsin(x\sqrt{1-y^2}+y\sqrt{1-x^2}) &\mbox{if } -\frac\pi2\le\arcsin x+\arcsin y\le\frac\pi2 \\ \pi- \arcsin(x\sqrt{1-y^2}+y\sqrt{1-x^2})& \mbox{if }\frac\pi2<\arcsin x+\arcsin y\le\pi \end{cases} $
and as like other trigonometric ratios are $\ge0$ for the angles in $\left[0,\frac\pi2\right]$
So, $\displaystyle\arcsin z\begin{cases}\text{lies in } \left[0,\frac\pi2\right] &\mbox{if } z\ge0 \\ \text{lies in } \left[-\frac\pi2,0\right] & \mbox{if } z<0 \end{cases} $
Case $(i):$ Observe that if $\displaystyle x\cdot y<0\ \ \ \ (1)$ i.e., $x,y$ are of opposite sign, $\displaystyle -\frac\pi2\le\arcsin x+\arcsin y\le\frac\pi2$
Case $(ii):$ If $x>0,y>0$ $\displaystyle \arcsin x+\arcsin y$ will be $\displaystyle \le\frac\pi2$ according as $\displaystyle \arcsin x\le\frac\pi2-\arcsin y$
But as $\displaystyle\arcsin y+\arccos y=\frac\pi2,$ we need $\displaystyle \arcsin x\le\arccos y$
Again as the principal value of inverse cosine ratio lies in $\in[0,\pi],$ $\displaystyle\arccos y=\arcsin(+\sqrt{1-y^2})\implies \arcsin x\le\arcsin\sqrt{1-y^2}$
Now as sine ratio is increasing in $\displaystyle \left[0,\frac\pi2\right],$ we need $\displaystyle x\le\sqrt{1-y^2}\iff x^2\le1-y^2$ as $x,y>0$
$\displaystyle\implies x^2+y^2\le1 \ \ \ \ (2)$
So, $(1),(2)$ are the required condition for $\displaystyle \arcsin x+\arcsin y\le\frac\pi2$
Case $(iii):$
Now as $\displaystyle-\frac\pi2\arcsin(-u)\le\frac\pi2 \iff -\frac\pi2\arcsin(u)\le\frac\pi2$
$\arcsin(-u)=-\arcsin u$
Use this fact to find the similar condition when $x<0,y<0$ setting $x=-X,y=-Y$