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?
On
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,
$$a=\sin^mx,x=\arcsin\sqrt[m]a ,$$
$$d(a)\propto\frac{\sqrt[m]a}{ma\sqrt{1-\sqrt[m]{a^2}}}.$$
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 $n\in[1,10^4]$:
$\hspace{2cm}$
Yves Daoust's answer deals with the case $\sin(X)^m$, too.