Why two characterizations of the arcsine distribution are equiv?

One way to prove this trigonometric identity is to $(1)$ first show that the derivatives of the functions on both sides of the equality are the same, and $(2)$ show that the two sides are equal when $x=0.$ \begin{align} & \frac d {dx}\,\frac 2 \pi \, \arcsin\sqrt x = \frac 2 \pi\cdot\frac 1 {\sqrt{1 - x}} \cdot \frac d {dx} \sqrt x = \frac 2 \pi \cdot \frac 1 {\sqrt{1-x}} \cdot \frac 1 {2\sqrt x} = \frac 1 \pi \cdot \frac 1 {\sqrt{x - x^2}}. \\[12pt] & \frac d {dx} \, \frac 1 \pi \arcsin(2x-1) = \frac 1 \pi \cdot \frac 1 {\sqrt{1- (2x-1)^2}} \cdot 2 = \frac 2 \pi \cdot \frac 1 {\sqrt{4x-4x^2}} = \frac 1 \pi \cdot \frac 1 {\sqrt{x-x^2}}. \end{align} (To make them equal when $x=0,$ one must of course add $\dfrac 1 2$ to the second one; that is omitted above because it's not involved in finding the derivative.)

A more conventional way (without calculus):

\begin{align} \text{Let } u & =\arcsin\sqrt x. \\[10pt] \text{Then } \sin u & = \sqrt x \\[10pt] x & = \sin^2 u \\[10pt] 2x-1 & = 2\sin^2 u - 1 \\ & = -\cos(2u) \text{ (This is the double-angle formula for the cosine.)} \\[10pt] \arcsin(2x-1) & = \arcsin(-\cos(2u)) = - \arcsin(\cos(2u)) \\[10pt] & = \arccos(\cos(2u)) - \frac \pi 2 = 2u - \frac \pi 2 = (2\arcsin\sqrt x) - \frac \pi 2. \end{align}

(That $\arccos(\cos(2u)) = 2u$ relies on the fact that $2u$ is between $0$ and $\pi$.)

Put $x^2$ for $x$ and use $\cos 2x = \cos^2x-\sin^2x =1-2\sin^2x$ and $\arcsin x+\arccos x =\pi/2$.

Multiply by $\pi$ and take the sine:

$$\sin(2\arcsin\sqrt x)=\sin(\arcsin(2x-1)+\frac\pi2)=\cos(\arcsin(2x-1)).$$

Then

$$2\sqrt x\sqrt{1-x}=\sqrt{1-(2x-1)^2},$$ and after squaring,

$$4x-4x^2=4x-4x^2.$$