is arcsin() evenly distributed?

Question

Given a uniform random distribution P of real numbers from [0,1] how might I prove (or disprove) that the map from P to Q of $(p\in{P} \rightarrow q=arcsin(p)\in{Q}) $ is a uniform distribution over [0,$\pi$]?

Answer 1

In a uniform distribution, the probability that a points lands on a interval depends only on the length of the interval.

Consider the intervals $A=[0,h]$ and $B=[\frac{\pi}{2}-h,\frac{\pi}{2}]$, which have the same length $h>0$.

The probability that $p$ lands on $A$ is higher than the probability that $p$ lands on $B$, because $\sin(A)$ has length $\sin(h)$ but $\sin(B)$ has length $1-\cos(h)$ and $\sin(h) > 1-\cos(h)$ for $0 < h < \frac{\pi}{2}$.