PDF of $\sin(2\pi X)$ where $X$ is uniform random variable

Question

Let $X$ be a continuous random variable with uniform distribution between $0$ and $1$. Compute the distribution of $Y = \sin(2\pi X)$.

$\sin(2\pi \cdot0)$ and $\sin(2\pi \cdot1) =0$. So, the inverse image of the function has multiple roots. How can I find the PDF of $Y$ then?

Answer 1

By symmetry, we can restrict $X\sim U(-0.25,0.25)$, since $X$ is uniform and the function sending it to $Y$ reflects on any half-integer multiple of 0.5. Then the inverse is well-defined: $$Y=\sin 2\pi X\implies X=\frac{\sin^{-1}Y}{2\pi}$$ $$J=\frac{\partial X}{\partial Y}=\frac1{2\pi\sqrt{1-Y^2}}$$ $$f_Y(y)=\frac1{\pi\sqrt{1-y^2}}\qquad y\in(-1,1)$$ where the multiplication by 2 in the last step is because the new $X$ has range 0.5. $Y$ is a scaled and shifted version of the arcsine distribution (or the arcsine distribution itself, depending on convention).

Answer 2

Since $X$ lies between $0$ and $1$ you can explicitly know what the two values of $X$ are if $Y = y \in [-1, 1]$ and hence, determine the probability of $Y \in [-1, y].$