Distribution of $Y = \sin X$ when $X$ is normal? Assume $X$ is Normally distributed :  $X\sim N(\mu,\sigma)$
What is the distribution of $Y = \sin X$ ?
I think we should start with $F_Y(y)=P(\sin X < y)$.
But how to continue?
 A: This is not an answer per se, but rather a reply to @user64494 who posted a solution (see comments above) using Maple code. Unfortunately, this reply is too long to fit into the comments section above, so I am using the answer box, if that is OK. 
User64494 states the solution is (using Maple):
 with(Statistics):X:=RandomVariable(Normal(mu,sigma)):Y:=sin(X):PDF(Y,t);

which produced the solution for the pdf $f(t)$:
$$f=\frac{\sqrt{2} \exp \left(-\frac{\left(\sin ^{-1}(t)-\mu \right)^2}{2 \sigma ^2}\right)}{\left(\sqrt{\pi } \sigma \right) \sqrt{1-t^2}}$$
for $-1<t<1$. I was surprised to see a closed-form solution posted, since ... to the best of my knowledge ... a closed-form solution to the infinite sum required is not known to exist. 
Monte Carlo check
To see that the posted Maple solution $f(t)$ is incorrect, simply compare a plot of the Maple solution $f(t)$ to the empirical pdf of $sin(X)$ when $X$~N($\mu$, $\sigma^2$), here with $\mu$ = $\pi$ and $\sigma = 2$:

In the above, the red curve is the Maple solution $f(t)$, and the blue curve is the empirical pdf. It appears clear that the posted Maple solution is in error. 
(I'd be happy to post the code to generate the plot if anyone is interested.)
More pertinently, the posted Maple pdf solution $f(t)$ does not integrate to unity over the domain of support.
A: I hope I understood you correctly, then it should be possible to consider $\cos (x) = \sin(x+\pi/2)$ and there are different appraoches here>>>, including calculation of the moments.
Generally it would be absolutely great if one could establish a theorem that considers both $\cos$ and $\sin$ then via taking $e^{ix}=\cos(x)+i\sin (x)$. Indeed there is hardly investigation made on this yet.
Hope this helps.
