How to prove this arcsine identity?
By Sebastian Wright
In the top of this Wikipedia article for the arcsine distribution it states that
$$\frac{2}{\pi} \arcsin(\sqrt{x}) = \frac{\arcsin(2x-1)}{\pi} + \frac{1}{2}$$
Why is this true? I haven't been able to derive this.
$\endgroup$ 32 Answers
$\begingroup$Here's one way: $$ \frac{2}{\pi} \arcsin(\sqrt{x}) \overset{\text{??}}{=} \frac{\arcsin(2x-1)}{\pi} + \frac{1}{2} $$
Differentiate both sides: $$ \frac2\pi\cdot\frac{1/(2\sqrt{x})}{\sqrt{1-x}} \overset{\text{??}}{=} \frac1\pi \frac{2}{\sqrt{1-(2x-1)^2}} $$
These are routinely seen to be equal, so the expressions above them must differ by a constant. Plugging in $x=0$ then finishes it off.
One could also just use trigonometric identities. Maybe I'll post more later....
$\endgroup$ 1 $\begingroup$Try this (up to domains)...
Let $\theta = \arcsin\sqrt{x}$ and $\varphi = \arcsin(2x-1)$. Therefore $\sin \theta = \sqrt{x}$ so $\sin^2\theta = x$. Also $\sin \varphi = 2x-1 =2\sin^2\theta-1=-\cos(2\theta) = \sin(2\theta-\pi/2)$. Therefore $\varphi = 2\theta - \pi/2$ so $\varphi+\pi/2 = 2\theta$ so $$ \frac{\varphi}{\pi}+\frac{1}{2} = \frac{2}{\pi}\theta, $$ which was to be proved.
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