Given that $f_\Phi(\phi)=\dfrac{\sin(\phi)}{2}$ for $\Phi\in[0,\pi]$, and $f_\Theta(\theta)=\dfrac{1}{2\pi}$ for $\Theta\in[0,2\pi]$, where $\Phi$ and $\Theta$ are independent. What is the PDF of $X=\sin(\Phi)\cos(\Theta)$?
So far, I have managed to obtain the PDFs of $U=\sin(\Phi)$ and $W=\cos(\Theta)$ as $$f_U(u)=\frac{u}{\sqrt{1-u^2}}, \quad\text{and}\quad f_W(w)=\frac{1}{\pi\sqrt{1-w^2}}.$$
I have attempted to follow the product distribution as described here. I know the answer should be $f_X(x)=\frac{1}{2}$, but I have been unable to progress further. How should I proceed?
The easiest solution returns to the sphere's geometry. The part of the surface ranging from $x$ to $x+dx$ is a ribbon of radius $\sqrt{1-x^2}$ and thickness $d\arcsin x=\frac{dx}{\sqrt{1-x^2}}$, so its area is $2\pi dx$. This can be used to prove the classical fact that a sphere's area matches that of a cylinder just long and wide enough to hold it, because each ribbon on the sphere's surface has the same area as its shadow on the cylinder. 3Blue1Brown has a great video on it here.