The sequence sin(n) does not converge

Suppose $\sin n\to\ell$. Then $\sin(n+1)\to\ell$ and $\sin(n-1)\to\ell$, so
$$2\sin1\cos n=\sin(n+1)-\sin(n-1)\longrightarrow0.$$
Since $\sin1\ne0$, we get $\cos n\to0$.

Two consequences.
- From $\sin^2n+\cos^2n=1$ we get $\ell^2=1$.
- The sequence $\sin2n=2\sin n\cos n$ tends to $2\ell\cdot0=0$. But $(\sin2n)$ is a subsequence of $(\sin n)$, so it tends to $\ell$. Hence $\ell=0$.

Contradiction. $\ell^2=1$ and $\ell=0$ cannot both hold. So $(\sin n)$ does not converge.

Remark. In fact, the set $\{n\bmod2\pi\}$ is dense in $[0,2\pi]$ and every value in $[-1,1]$ is a limit point of $\sin n$.