End behavior of $(\sin k)^k$ where $k \in \mathbb{N}$ It's clear that the set $\{\sin k \mid k \in \mathbb{N}\}$ is dense in $[0, 1)$. What I'm having trouble determining is how the sequence $\{(\sin k)^k \mid k \in \mathbb{N}\}$ behaves for large $k$.
My main question is: does there exist $N \in \mathbb{N}$ and $\epsilon \in \mathbb{R}, 0 \leq \epsilon < 1$ such that for all $k \geq N, k \in \mathbb{N}$, $(\sin k)^k \leq \epsilon$? Or equivalently, does there exist $N \in \mathbb{N}$ such that $\sup(\{(\sin k)^k \mid k \in \mathbb{N}, k \geq N\}) \neq 1$?
 A: So I think this is actually an open question, because it's so closely related to the irrationality measure of $\pi$. We are essentially asking whether $k$ can be really close to a number of the form $n\pi + \pi/2$, because that's a necessary condition to have $\sin k \approx 1$. In fact, we need to have $\sin k \approx 1$ to an error that is controlled even after raising to power $k$. This is pretty delicate.
We can write down rational approximations of $\pi/2$. In particular, suppose we can find a sequence of rational numbers such that
$$\left|\frac{\pi}{2} - \frac{p_j}{q_j}\right| \le \frac{1}{q_j^{2 + \mu}}.$$
for some $\mu > 0$ (equivalently, that the irrationality measure of $\pi$ is strictly greater than $2$).
Without loss of generality, $q_j$ is odd. In that case, we have
$$\left|p_j - q_j \frac{\pi}{2}\right| \le \frac{1}{q_j^{1 + \mu}} \implies \left|p_j - (2n_j + 1)\frac{\pi}{2}\right| \le \frac{1}{q_j^{1 + \mu}}$$
for some integer $n_j$. Now $p_j$ is our candidate index; we have
$$1 - |\sin p_j|^{p_j} \le 1 - \left(1 - \frac 1 {q_j^{1 + \mu}}\right)^{p_j} \approx \frac{p_j}{q_j^{1 + \mu}} \approx \frac{1}{q_j^{\mu}} \to 0$$
as $j \to \infty$. In this case, the supremum of $\{(\sin p_j)^{p_j}\}$ is equal to $1$.

On the other hand, suppose that there were an $\epsilon > 0$ such that for all sufficiently large $p, q$ we had
$$\left|\frac{\pi}{2} - \frac p q\right| \ge \epsilon \frac{1}{q^2}.$$
Then we can play the same game, finding that
$$1 - |\sin p|^p$$
is bounded below, and so the supremum is less than $1$.
