Show that the set of accumulation points of the sequence $a_n = \sin{n}$ is the closed interval $[-1, 1]$ I'm learning about Real Analysis, specifically accumulation points of a sequence, and need some help with this problem: 

Show that the set of accumulation points of the sequence $a_n = \sin{n}$ is the closed interval $[-1, 1]$.


My thoughts: 
Obviously we have $-1 \leq \sin{n} \leq 1 \forall n \in \mathbb{N}$. 
The set $a_n$ will be of the form 
$a_n = \{\sin1, \sin2, \sin3, ...\}$. 
Given $l \in [-1, 1]$ we want to show that there exists a subsequence 
$\{a_{n_k}\}$ such that $a_{n_k} \rightarrow l$ as $k \rightarrow \infty$.

This problem is different from the problems I'm used to solve in the sense that instead of finding the accumulation points of a sequence I need to prove that the accumulation points of a sequence is a closed interval and I don't know how to approach this. 
 A: Consider the bounded sequence $$n\mapsto z_n:=e^{in}\in{\mathbb C}\qquad(n\in{\mathbb N}_{\geq0})\ .$$ 
By Bolzano's theorem the $z_n$ have an accumulation point $\zeta\in S^1$. It follows that for any $\epsilon>0$ there are natural numbers $p<q$ with $|z_q-z_p|<\epsilon$. Put $r:=q-p$, and consider the subsequence $$k\mapsto w_k:=z_{k r}=e^{i kr}\quad(k\geq0)\ .$$ From
$w_{k+1}-w_k=e^{ik r}(e^{ir}-1)$
it follows that $|w_{k+1}-w_k|=|e^{iq}-e^{ip}|$; hence consecutive $w_k$ have a distance $<\epsilon$. From this we can conclude that the set  $Z:=\{z_n\>|\>n\geq0\}$ is in fact dense on $S^1$. 
Now let an arbitrary $y\in[{-1},1]$, as well as an $\epsilon>0$, be given. There is an $x\geq0$ with $z:=(x,y)\in S^1$, and there is a $z_n\in Z$ with $|z-z_n|<\epsilon$. It follows that
$$|y-\sin n|\leq|z-e^{in}|=|z-z_n|<\epsilon\ .$$
As $\epsilon>0$ was arbitrary we can conclude that $y$ is an accumulation point of the sequence $(\sin n)_{n\geq0}$.
A: Hint: This is equivalent to show that the set of rest of the division of $n\in N$ by $2\pi$ is dense in $[0,2\pi]$. This follows from the fact that the subgroup generated by $n\in N$ and $2\pi$ is dense in $R$.
