# Equidistribution of $\sin(n)$

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?

• show that $n / (2 \pi) - \lfloor n / (2 \pi) \rfloor$ is equidistributed, and consider for $x \in \ ]-1,1[$ : $\displaystyle f_\epsilon(x) = \lim_{N \to \infty} \frac{\# \{ \ |\sin(n) - x| < \epsilon \ \mid \ 1 \le n \le N\}}{N}$, you should get that when $\epsilon \to 0$, $f_\epsilon(x) \to C |\sin(2 \pi x)|$ with $C = \int_0^1 |\sin(2 \pi x)| dx$. – reuns May 6 '16 at 16:45
• One should not expect equidistribution in this case: the distribution will be "denser" near $1$ and $-1$. We can derive the exact result using equidistribution on the circle and considering the $y$-coordinate. – Ben Grossmann May 6 '16 at 16:51
• @user1952009 Your computations seem to be wrong. – Did May 6 '16 at 18:02
• @Did yes I saw that, I need to use $|arcsin'(x)|$ for computing $f_\epsilon(x)$ – reuns May 6 '16 at 18:47
• I, likewise, didn't know how to add an image to an answer back then. I have added a plot comparing the computed and predicted pdfs. – John Barber May 11 '20 at 3:58

## 1 Answer

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 $$1\leq n\leq {10}^6$$.