Suppose $\Theta \sim Unif(0,2\pi)$ and $s \in (0,1)$. I am trying to find $P(\mid \sin(\theta) \mid < \sqrt s)$.
My attempt:
We have that
$$P(\mid \sin(\theta) \mid < \sqrt s) = P(-\sqrt s < \sin(\theta) < \sqrt s) = \\
= P(\sin(\theta) < \sqrt s) – P(\sin(\theta) < – \sqrt s) = \\ =P(\theta < \arcsin(\sqrt s)) – P(\theta < \arcsin(-\sqrt s)) = \\
= \frac{\arcsin(\sqrt s)}{2\pi} – \frac{\arcsin(-\sqrt s)}{2\pi} = \frac{\arcsin(\sqrt s)}{\pi}$$
where I've used the oddity of the arcsine function.
However, taking the derivative with respect to $s$ of this laxt expression yields
$$\frac{d}{ds}\frac{\arcsin(\sqrt s)}{\pi} = \frac{1}{\pi\sqrt{ s(1-s)}}ds$$
which integrates to $\frac{1}{2}$. Where did I miss a $2$ along the way?
Best Answer
For $x\in [0,1]$, $$ \mathsf{P}(\sin(\theta)\le -x)=\mathsf{P}(\sin(\theta)\ge x)=\mathsf{P}(\arcsin(x)\le \theta\le \pi-\arcsin(x)). $$ Thus, $$ \mathsf{P}(|\sin(\theta)|\le x)=1-2\mathsf{P}(\arcsin(x)\le \theta\le \pi-\arcsin(x))=\frac{2\arcsin(x)}{\pi}. $$