There are a number of ways of show this. Here is an elementary way: Suppose on the contrary that
\(\sin(n) \to L\text{.}\) Then both
\(\sin(n+1)\) and
\(\sin(n-1)\) converge to
\(L\) as well. Then use the identities:
\begin{gather*}
\sin(n+1) + \sin(n-1) \equiv 2\sin(n)\cos(1)\\
\sin(n+1) - \sin(n-1) \equiv 2\sin(1)\cos(n)
\end{gather*}
to conclude that
\(\sin(n) \to 0\) and that
\(\cos(n)\to 0\) as well. This gives us a contradiction as
\(\sin^2(n) + \cos^2(n) \equiv 1\text{.}\)