Exam exercise on sequence $a_n = \sin(n)$

Question

Prove that the sequence $a_n = \sin(n)$ cannot converge when $n \rightarrow \infty $

I tried to find two subsequences that converge to different values but I am having trouble with the fact that $n \in N$ and so I can't pick values in the whole domain of the function.

Maybe there is an other way.

Answer 1

If the sequence converges, then so does $\cos(n)$. Say limit of $a_n$ is $L$. Then limit of $\cos(n)$ is $\sqrt{1-L^2}$.

Now, if a_n converges then so does $a_{2n}$ and to the same value $L$.

But $\sin(2n)=2\sin(n)\cos(n)$. Taking limits you can get the value of $L$.

Now, fix $m$ and consider the subsequence $a_{n+m}$. This also converges to $L$.

We also have $\sin(n+m)=\sin(n)cos(m)+\cos(n)\sin(m)$. Taking limits as $n \rightarrow \infty$ we get $L=L\cos(m)+\sqrt{1-L^2}\sin(m)$. Now derive a contradiction.

(You would also need the fact that if $L$ exists, it is non zero by considering large $n$ where $\sin(n)$ is non zero and bounded away from $0$)