Suppose $X$ is a random variable drawn from a normal distribution with mean $E$ and variance $V$. How could I calculate variance of $\sin(X)$ and $\cos(X)$?
(I thought the question was simple and tried to do a search, but did not find any good answer.)
What if there is no assumption about the distribution of $X$, and only sample mean and variance are provided?
Best Answer
What is below is for $\mu=0$ (and variance renamed $\sigma^2$). Then $\mathbb{E}[\sin X]=0$, and you have $$ \operatorname{Var} \sin X = \mathbb{E}[\sin^2 X] = \frac{1}{2}\left(1-\mathbb{E}[\cos 2X]\right) $$ and $$ \mathbb{E}[\cos 2X] = \sum_{k=0}^\infty (-1)^k\frac{2^{2k}}{(2k)!} \mathbb{E}[X^{2k}] = \sum_{k=0}^\infty (-1)^k\frac{2^{2k}}{(2k)!} \sigma^{2k} (2k-1)!! = \sum_{k=0}^\infty (-1)^k \frac{2^{k}\sigma^{2k}}{k!} = e^{-2\sigma^{2}} $$ and therefore $$ \operatorname{Var} \sin X = \boxed{\frac{1-e^{-2\sigma^2}}{2}} $$ You can deal with the variance of $\cos X$ in a similar fashion (but you now have to substract a non-zero $\mathbb{E}[\cos X]^2$), especially recalling that $\mathbb{E}[\cos^2 X] = 1- \mathbb{E}[\sin^2 X]$.
Now, for non-zero mean $\mu$, you have $$ \sin(X-\mu) = \sin X\cos \mu - \cos X\sin\mu $$ (and similarly for $\cos(X-\mu)$) Since $X-\mu$ is a zero-mean Gaussian with variance $\sigma^2$, we have computed the mean and variance of $\sin(X-\mu)$, $\cos(X-\mu)$ already. You can use this with the above trigonometric identities to find those of $\cos X$ and $\sin X$. (it's a bit cumbersome, but not too hard.)
Without knowing anything about the distribution of $X$, I don't think there's much you can do.