In this post I have proved the result that \[
\overline{\sin\mathbb{Z}}=\overline{\cos\mathbb{N}}=[-1,1]
\] by purely group theoretical approach. Since I am going to make tutorial notes for the Real Analysis class next semester, I rethink about this problem and further improve the result such that \[
\overline{\sin\mathbb{N}}=[-1,1],
\] based on the findings in that post.
Now by the density we can find intergers $ n_k$ such that $ \sin n_k\to 0$. Let $ p_{k}=|n_k|$, then $ \mathop{\mathrm{sgn}}(n_k) \sin n_k = \sin p_k\to 0$. Pick an $ a\in [-1,1]$, we now show that $a\in \sin \mathbb N$. First of all, there is a sequence of integers $\{h_k\}$ such that $\sin h_k\to a$. Recall the identity that \[\sin x - \sin y = 2\cos \frac{x+y}{2}\sin \frac{x-y}{2},\] we have for each $ k$, \[
\sin (h_k+2p_i)-\sin h_k=2\cos (h_k+p_i)\sin p_i\to 0 \quad i\to \infty,\] meaning that $\displaystyle\lim_{i\to\infty}\sin (h_k+2p_i) = \sin h_k$.
Now we are almost done, for each $n\in \mathbb N$, we can find an $ k_n$ such that $ |\sin h_{k_n} -a|<1/n$. Fix this $ k_n$, by the last limit we obtained, we can find an index $ i_n$ such that $ |\sin (h_{k_n}+2p_{i_n})-\sin h_{k_n}|<1/n$ and $ P_n:= h_{k_n}+2p_{i_n} > 0$. Hence \[|\sin P_n-a|\leq |\sin P_n-\sin h_{k_n}|+|\sin h_{k_n} - a|<2/n,\] we conclude $ a\in \overline{\sin \mathbb N}$.
Consequence: The subsequential limits of $ \{\sin n:n\in \mathbb N\}$ are precisely those in $ [-1,1]$.
