Computing a Inf-Sup of $\sin(nx)$ I would like to compute the following expression
$$\inf_{x\in ]0,\pi[}\Big(\sup_{n\in \Bbb Z}|\sin(nx)|\Big).$$
Any Hint or answer is warmly appreciated.
 A: $$\inf_{x\in\left]0,\pi\right[}\left(\sup_{n\in\mathbb{Z}}\vert\sin(nx)\vert\right)=\frac{\sqrt{3}}{2}$$
Proof:
As has been pointed out by @Crostul, we have $\limsup_{n\to\infty}\vert\sin(nr\pi)\vert=1$ if $r\not\in\mathbb{Q}$.
Therefore, it suffices to consider $x\in]0,\pi[\cap\pi\mathbb{Q}$.
Hence, assume $x=\frac{m}{k}\pi$ with $m,k\in\mathbb{N}$, $m<k$, $m$ and $k$ are relatively prime.
Case 1: $k$ is even.
Choose $n:=\frac{k}{2}$. Then, $nx=\frac{m}{2}\pi$. Since $m$ is odd, we have $\vert\sin(nx)\vert=1$.
Case 2: $k$ is odd.
Let $k=p_1...p_s$ where all the $p_i$ are prime. Define
$$n_1:=p_1...p_{s-1}$$
and
$$n_2\in\mathbb{N}\text{ such that }mn_2\equiv\frac{p_s+1}{2}\ (\mathrm{mod}\ p_s).$$
Note that this is possible since $p_s\neq 2$ and is prime. Now, let $n:=n_1n_2$.
Then, $nx=\frac{m}{k}n_1n_2\pi=\frac{mn_2}{p_s}\pi=\left(z+\frac{1}{2}+\frac{1}{2p_s}\right)\pi$ for some $z\in\mathbb{N}$. Thus, the distance of $nx$ from a point where $\sin$ has absolute value $1$ is at most $\frac{1}{2p_s}\leq\frac{1}{6}$. This means that $\vert\sin(nx)\vert\geq\frac{\sqrt{3}}{2}$.
Finally, choosing $x:=\frac{\pi}{3}$, we find
$$\sup_{n\in\mathbb{Z}}\left\vert\sin\left(\frac{n\pi}{3}\right)\right\vert=\frac{\sqrt{3}}{2}.$$
