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Is the sequence $a_{n}=\sin(n)$ uniformly distributed on the interval $[-1,1]$?

For the first $10000$ $n$, it seems as if it is more dense at $-1$ and $1$ than in the middle.

Is there any way to prove this or its uniform distribution?

Also, can a result be generalized for $\sin^m(n)$ over its appropriate interval?

Stephen
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  • Relevant: http://math.stackexchange.com/questions/4764/sine-function-dense-in-1-1 and http://math.stackexchange.com/questions/1084945/is-sin-nm-mid-n-in-mathbbn-dense-in-1-1-for-every-natural-nu. (For density, not uniform distribution) – Clement C. Mar 02 '17 at 16:56

2 Answers2

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By Weyl's equidistribution theorem, the sequence $\{e^{in}\}_{n\geq 1}$ is uniformly distributed over the unit circle, by the irrationality of $\pi$. The distribution of $\sin(n)$ over $(-1,1)$ is so the same as the distribution of $\sin(X)$, where $X$ is a random variable uniformly distributed over $(0,2\pi)$.
The PDF of such distribution is so given by $\frac{1}{\pi\sqrt{1-t^2}}$, supported on $(-1,1)$.
Here it is a histogram representing the distribution of $\sin(n)$ for $n\in[1,10^4]$:

$\hspace{2cm}$enter image description here

Yves Daoust's answer deals with the case $\sin(X)^m$, too.

Jack D'Aurizio
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If the points $n\bmod 2\pi$ are uniformly distributed, you can expect a density which is proportional to the inverse of the slope of the curve, i.e.

$$a=\sin x, x=\arcsin a,$$ $$d(a)\propto\frac1{\sqrt{1-a^2}}.$$

For the case with the exponent,

$$a=\sin^mx,x=\arcsin\sqrt[m]a ,$$

$$d(a)\propto\frac{\sqrt[m]a}{ma\sqrt{1-\sqrt[m]{a^2}}}.$$