This is a claim that I had made while finding the supremum of $(\sin(n))_{n \in \Bbb{N}}$. The supremum is $1$ if there exists $n \in \Bbb{N}$ such that for all $\epsilon' > 0$, $1-\epsilon' < \sin(n)$. For this to happen, the following claim must be valid:
There exists $n, m \in \Bbb{N}$ such that $$2m\pi + \frac{\pi}{2} - \epsilon < n < 2m\pi + \frac{\pi}{2}$$
This intuitively feels correct, and I came up with a (shaky) proof for this claim. I tried to use the density theorem as follows:
By the density theorem, there exists $u, v, k \in \Bbb{N}$ such that $$2k\pi + \frac{\pi}{2v} - \frac \epsilon v < \frac uv < 2k\pi + \frac{\pi}{2v}$$ Multiplying the inequality by $v$, we see that $$2kv\pi + \frac{\pi}{2} - \epsilon < u < 2kv\pi + \frac{\pi}{2}$$ if we let $m = kv$, then $u$ satisfies the claim. $\blacksquare$
However, I don't think this is valid. Along similar lines, consider $t + \frac{5}{4v} < \frac{u}{v} < t + \frac{6}{4v}$. Multiplying by $v$, we get $tv + \frac{5}{4} < u < tv + \frac{6}{4}$, which does not hold for any $t, v, u \in \Bbb{N}$.
Could anyone suggest a better proof for this claim? Any hints or strategies are appreciated as well.
