Continue the work before

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 $[-1,1]$.