Show the $\arcsin$ identity: $ \arcsin(1 - 2x) + 2\arcsin(\sqrt{x}) = \pi / 2$

Let $f(x)=\arcsin ( 1 - 2x) + 2 \arcsin\sqrt x$ where $x\in[0,1]$ so that \begin{align}f'(x)&=\frac1{\sqrt{1-(1-2x)^2}}\cdot(-2)+2\cdot\frac1{\sqrt{1-\sqrt x^2}}\cdot\frac1{2\sqrt x}\\&=-\frac2{\sqrt{4x-4x^2}}+\frac1{\sqrt{x(1-x)}}\\&=-\frac1{\sqrt{x(1-x)}}+\frac1{\sqrt{x(1-x)}}=0\end{align} for all $x\in(0,1)$. Hence in this interval, we have that $f$ is constant. Since $f(0)=f(1/2)=f(1)=\pi/2$, we conclude that $$\arcsin ( 1 - 2x) + 2 \arcsin\sqrt x=\frac\pi2.$$

Here’s an elementary proof. Let $x=\sin^2\phi$. Then, since $\cos(2\phi)=1-2\sin^2\phi$, we have $$\arcsin(1-2x)=\arcsin(1-2\sin^2\phi)=\arcsin(\cos(2\phi))=\frac{\pi}{2}-2\phi$$ and $$2\arcsin(\sqrt{x})=2\arcsin(\sin\phi)=2\phi$$ From which the result follows elementarily (without calc).