A subsequence of $\{|\sin n|\}$ that converges to $0$ "fast" It is well-known that for each $a\in [0,1]$ there is a subsequence of $\{|\sin n|\}_{n=0}^\infty$ that converges to $a$. I am curious about the speed of convergence. In particular, for each $\alpha>0$, is there a subsequence $\{|\sin n_k|\}_{k=0}^\infty$ which converges to $0$ "fast" in the sense that
$$|\sin n_k|\le\frac{1}{n_k^\alpha},\,\forall\,k\in\mathbf N\,?$$
If not, what restriction should be imposed on $\alpha$?
Any hints or reference (to existing literature on it) will be highly appreciated.
 A: The exact irrationality measure of $\pi$ is still unknown, it is conjectured to be $2$ but at the moment we only know that
$$ \left|\pi-\frac{p}{q}\right|\leq \frac{1}{q^{7.11}} $$
holds for a finite number of fractions $\frac{p}{q}$, while
$$ \left|\pi-\frac{p}{q}\right|\leq \frac{1}{q^2} $$
certainly holds infinite times due to the irrationality of $\pi$. So if $\frac{p}{q}$ is a convergent of the continued fraction of $\pi$ we have
$$ \left|\pi q-p\right|\leq\frac{1}{q} $$
and since $\sin(x)$ is a Lipschitz-continuous function
$$ \left|\sin p\right| \leq \frac{1}{q} \approx \frac{\pi}{p} $$
i.e. your inequality is achieved by $\alpha=1-\varepsilon$. $1$ is probably the supremum of the set of $\alpha$ allowing the inequality to hold, but the truth of this assertion is beyond our current technology. Anyway, for the unknown optimal $\alpha\in[1,6.11]$ we have that $\{n_k\}_{k\geq 1}$ is a subsequence of
$$ 3,22,333,355,103993,104348,208341,312689,833719,1146408,\ldots $$
given by the numerators of the convergents.
