If $\theta\sim Unif[-1,1]$, then what is the CDF of $U=\sin(\pi\theta)$?
Now, its easy to see that $$P_{U}(t) = P\left(\theta \leq\frac{\sin^{-1}(t)}{\pi}\right)$$
somehow the answer is equal to :
$$ \mathrm{if} \,\,\,\, \, 0 \leq t \leq 1 : P(U\leq t) = \frac{1+\frac{2 \,\sin^{-1}(t)}{\pi}}{1-(-1)}$$
similarly for the other part of the interval but with: $$\frac{1-\frac{2 \,\sin^{-1}(-t)}{\pi}}{1-(-1)}$$
The denominator is of course form the definition of the Unifrom dist as the length of the interval. But I cant get how we obtained that numerator value, why is there a $1-2..$ I tried drawing the graph but I cant get it still.
Any hints about the clarification are appreciated!
To help to calculate the probability $$P(\sin(\pi \theta)\leq u)$$ take a look at the following figure:
In the upper figure, $u>0$ and if $\pi\theta$ falls in the thick red region then the event in question occurs. That is, for $1\geq u>0$ the probability (the value of the cdf) at $u$ is
$$\frac{\pi+2\sin^{-1}(u)}{2\pi}=\frac12+\frac{\sin^{-1}(u)}{\pi}.$$
In the lower figure, one can see when the event in question takes place; look at the thick blue line. So the probability (the value of the cdf) for $-1\leq u\leq0$ is
$$\frac{\pi-2|\sin^{-1}(u)|}{2\pi}=\frac12-\frac{|\sin^{-1}(u)|}{\pi}.$$