distribution of $\cos(\omega_0 n)$ where n are integers? Assume we have the sequence $\,x[n]=\cos(\omega_0 n),$ where $n$ are integers. 
If we suppose these are realizations of a random variable, what would be the p.d.f. of that random variable?
 A: So the point of this answer is not so much to answer the question posed as it is to give the question a proper formulation.  However, I find this might be a little too long for a comment, so here it goes.

So if $f(x)$ is the desired pdf, we'll say that 
$$F(x) = \int_{-\infty}^{x} f(x) = \int_{-1}^{x} f(x)
\Rightarrow f(x)=F'(x)$$
is the associated cdf.  The question we'd like to answer then is what is the closed form of $F(x)$, where $F(x)$ is the probability that $\cos\omega_0 n<x$, given that $\omega_0$ is not a rational multiple of $\pi$ and $n$ is uniformly selected from $\mathbb N$.

Here, then, is my attempt at an answer:

Conjecture: let 
  $C\subset\mathbb C = \{z\in\mathbb C:|z|=1\}$.
  Then e$^{i \omega_0 n}$ is uniformly distributed over $C$.

Assuming the above to be the case, we note:
$$
\cos\omega_0 n < x \Rightarrow \\
\text{Re}\{\exp(i\omega_0 n)\} < x \Rightarrow\\
\arccos(x) <\text{Arg}(\exp(i\omega_0 n))< 2\pi-\arccos(x)
$$
It follows that
$$
F(x)=\frac{2\pi-2\arccos(x)}{2\pi}=1-\frac1{\pi}\arccos(x)
$$
It follows that the desired pdf should be
$$
f(x) = 
\begin{cases}
\frac{1}{\pi\sqrt{1-x^2}} & -1\leq x\leq 1\\
0 & \text{otherwise}
\end{cases}
$$
