Variable transformation $Y= \sin (X)$ where $X$ is uniform on $[0, \pi]$ Given $X$ is uniform on $[0, \pi]$.
Let $Y= \sin (X)$ then $y \in [0,1].$
What is the distribution of Y? If intuition serves me correct I would think that Y is uniform on $[0, 1]$. But when I start to do the variable transformation I get stuck.
$P(Y \le y) = P(\sin (X) \le y) = P(X \le \arcsin(y) = \frac{arcsin(y)}{\pi}$ for $y \in [0,1].$
Then $ \arcsin(y)$ would take on two values for each $y$ value and this is where I am stuck.
 A: You were almost there.  Because, as you noticed, almost every point on the support of $Y$ maps to two points on the support of $X$, we must consider both those contributions. 
If we take $\;\color{navy}{\arcsin : [-1;1]\mapsto [-\pi/2; \pi/2]}$ , then the cumulative distribution function of $Y$ is:
$$\begin{align}
\mathsf P(0\leq Y\leq y) & = \mathsf P\Big(\big(0\leq X \leq \arcsin(y)\big) \cup \big(\pi-\arcsin(y)\leq X\leq \pi\big)\Big)
\\ & = \frac{\arcsin(y)+(\pi-\pi+\arcsin(y))}{\pi} \mathsf 1_{[0;1)}(y)+\mathsf 1_{[1;\infty)}(y)
\\ & = \frac{2\arcsin(y)}{\pi} \mathsf 1_{[0;1)}(y)+\mathsf 1_{[1;\infty)}(y)
\end{align}$$ 
Because if we plot $y=\sin(x)$ for $x\in[0;\pi]$ then we notice that there are two intervals on the support of $X$ where $Y\leq y$.   These correspond to the intervals $X\in [0;\arcsin(y)]$ and $X\in [\pi-\arcsin(y);\pi]$.
Now, can you find the probability density function?  (Note: This is not uniformly distributed.)
A: Confused? Draw the graph of $\sin x$, for $0\leq x\leq\pi$. Is it one-to-one? If the answer is no, then you have to tread carefully.  Draw a horizontal line at a fixed $y$ and look for all the $x$ values in $[0,\pi]$ that satisfy your inequality $\sin(x)\leq y$.  You'll realize they don't form an interval, but the union of two disjoint intervals (symmetric with respect to $x=\pi/2$), unless $y=1$.
Still confused? Learn about Iverson brackets (and thank Don Knuth).
