TITLE: $\sin(n)$ subsequence limits set
QUESTION [13 upvotes]: We were given a challenge by our calculus professor and I've been stuck on it for a while now. 
Show that the set of subsequence limits of $A_n=\sin(n)$ is $[-1, 1]$ (another way to phrase this would be: $\forall r\in [-1,1]$ there exists a subsequence of $A_n=\sin(n)$ that converges to $r$).
What would be a good way to start?

REPLY [2 votes]: Suppose you want to find positive integers $n, k$ such that $|n - 2 \pi k - \arcsin r| < \frac{1}{m}$ for some large positive integer $m$ and $r \in [-1, 1]$.  Then I claim that some value of $n$ less than $2\pi m$ works.  This is the pattern I was trying to get you to see, although I wasn't doing a very good job of it (I should've asked you to plot $k$ instead of $n$).  
Unfortunately I think this is a little hard to prove except when $r = 0$; fortunately, using the second half of Marek's argument you can prove the same result with a weaker bound on $n$ for general $r$ using the case of $r = 0$, so I would encourage you to try that case first.  To do this, try plotting the fractional parts of the numbers $\frac{n}{2\pi}$ in $[0, 1)$.