Tukukkk

Suppose $X$ is a random variable drawn from 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?

What si 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
$$
mathbb{E}[cos 2X] = sum_{n=0}^infty (-1)^kfrac{2^{2k}}{(2k)!} mathbb{E}[X^{2k}]
= sum_{n=0}^infty (-1)^kfrac{2^{2k}}{(2k)!} sigma^{2k} (2k-1)!!
= sum_{n=0}^infty (-1)^k frac{2^{k}sigma^{2k}}{k!} = e^{-2sigma^{2}}
$$
and therefore
$$
operatorname{Var} sin X = boxed{frac{1-e^{-2sigma^2}}{2}}
$$
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
$$
sin(X-mu) = sin Xcos mu - cos Xsinmu
$$
(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.

Here is a general formulation using the law of the unconscious statistician. that can be applied to other functions too. For specific calculations with $sin$ and $cos$ here though, I would say Clement C.'s answer is better!

The mean of $color{blue}{h(X)}$ (for some function $h$) would be given by the integral
$$mathbb{E}[h(X)]=int_{-infty}^{infty}color{blue}{h(x)}f_X(x), dx,$$
where $f_X$ is the probability density function of $X$.

The second moment would be found similarly as $$mathbb{E}left[(h(X))^2right] = int_{-infty}^{infty}color{blue}{(h(x)^2)}f_X(x), dx.$$

Once you know the first two moments here, you can calculate the variance using $mathrm{Var}(Z) = mathbb{E}[Z^2] - (mathbb{E}[Z])^2$.

Replace $h(x)$ with $cos x$ for the corresponding expectations for $cos X$, and similarly with $sin x$.

If the distribution of $X$ is not known, we cannot generally compute the exact mean and variance of $h(X)$. However, you may want to see this for some approximations that could be used. Some useful ones for you may be that if $X$ has mean $mu_X$ and variance $sigma^2_X$, then
$$mathbb{E}[h(X)]approx h(mu_X) + dfrac{h''(mu_X)}{2}sigma_X^2$$
and
$$mathrm{Var}(h(X))approx (h'(mu_X)^2)sigma^2_X + dfrac{1}{2}(h''(mu_X))^2 sigma^4_X.$$

$cos^2(x) = frac{cos(2x)+1}2$, which averages out to $frac12$. So as the variance of $X$ goes to infinity, the variance of $cos(X)$ goes to $frac12$, assuming the distribution of $X$ is "well-behaved". The lower bound is $0$ (the variance can be made arbitrarily small by choosing the variance of $X$ to be small enough), and as @angryavian says, the upper bound is $1$. Since $|cos(x)| leq 0$, and the inequality is strict for all but a measure zero set, the variance of $cos(X)$ is less than the variance of $X$.