Why two characterizations of the arcsine distribution are equiv? According to Wikipedia the CDF of arcsine dist. is:
$$F(x)=\frac2{\pi}\arcsin(\sqrt{x})=\frac{\arcsin(2x-1)}{\pi}+\frac12$$
So, why are these two equivalent?
Thanks in advance.
 A: 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 $\pi$.)
A: Put $x^2$ for $x$ and use $\cos 2x = \cos^2x-\sin^2x =1-2\sin^2x$
and $\arcsin x+\arccos x =\pi/2$.
A: 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,
$$4x-4x^2=4x-4x^2.$$
