I would like to find the PDF of the random variable $Y=\sin(X)$ given the PDF of $X$:
$$f(x) = \frac{2x}{\pi^2} \text{ for } 0<x<\pi \text{ and } 0 \text{ otherwise}$$
Following the tips in the question here: Transformation of PDF
I found $$f(y)= \frac{1}{\sqrt{1-y^2}}\frac{2\sin^{-1}(y)}{\pi^2}$$
However, I am not sure about the range of $Y$. Is it correct to say that $Y$ is between $1$ and $-1$ for the above $f(y)$?
Since $\sin\, x$ takes values between $0$ and $1$ when $x$ is between $0$ adn $\pi$, the range for $y$ is $(0,1)$, not $(-1,1)$. For a proper application of the transformation formula you have to split the range $(0, \pi)$ of $x$ into $(0, \pi /2)$ and $(\pi /2, \pi)$.