Equidistribution of $\sin(n)$

Question

It is a classical result that $$ \limsup_n \sin(n) = 1 $$ Even more, the set $\{\sin(n):n\in\mathbb{N} \}$ is dense in $[-1,1]$. I was wondering if it is possible to say something about the distribution of the sequence in the interval $[-1,1]$. I have no idea if one should expect equidistribution or not. Is there any result about this question?

Answer 1

Assuming $n \mod 2\pi$ is distributed uniformly on $[0,2\pi]$, we can model the distribution of $\sin(n)$ as having the same distribution as $\sin(\frac{\pi}{2}u)$, where $u$ is uniformly distributed on $[-1,1]$. $u$ thus has the probability density function (pdf) $$ f(u) = \frac{1}{2} $$ on $u\in [-1,1]$ and zero everywhere else. What we want is the pdf $g(x)$ of the quantity $x = \sin(\frac{\pi}{2}u)$. This is given by $$ g(x) = \left|\frac{du}{dx}\right| f(u(x)). $$ Since $u(x) = \frac{2}{\pi} \sin^{-1}(x)$ and $f = \frac{1}{2}$ in the region of interest, this becomes: $$ g(x) = \frac{1}{\pi\sqrt{1 - x^2}}. $$ I've checked this by looking at the distribution of $\sin(n)$ for the first 1,000,000 integers $n$, and I find that it matches this prediction perfectly.

Edited 4 years later to include a comparison plot:

Here the black curve is the function $g(x)$ given above, and the red dots are the computed pdf of $\sin(n)$ obtained by looking at $1\leq n\leq {10}^6$.