Prove the divergence of the sequence $\left\{ \sin(n) \right\}_{n=1}^{\infty}$. I am looking for nice ways of proving the divergence of the sequence $\left\{x_n\right\}_{n=1}^{\infty}$ defined by
$$x_n=\sin{(n)}.$$
One (not so nice) way is to construct two subsequences: one where the indexes are picked such that they lie in the intervals 
$$I_k=\left(\dfrac{\pi}{6}+2\pi(k-1),\dfrac{5\pi}{6}+2\pi(k-1)\right)$$
and one where they lie in
$$J_k=\left(\dfrac{7\pi}{6}+2\pi(k-1),\dfrac{11\pi}{6}+2\pi(k-1)\right).$$
If ${x_n}$ converges, then all its subsequences must converge to the same limit, but here the first subsequence has all its values in the interval $\left[ \frac{1}{2}, 1\right]$ while the second has its values in $\left[-1,-\frac{1}{2}\right]$. Contradiction.

Filling the details of this proof is rather tedious and not very elegant. Anyone has a better idea?
Equidistribution argument is very elegant, but it becomes a sledgehammer method when it comes to a question of mere convergence.

A simple argument can reveal the divergence of $(\sin n, n \geq 1)$. Let

$$ (x_n, y_n) = (\cos n, \sin n)$$

(I changed the notation for the sake of consistency of notation.) Then the application of the addition formula for trigonometric function or the rotation matrix gives

$$ \begin{align*}
x_{n+1} &amp;= x_n \cos 1 - y_n \sin 1 \\
y_{n+1} &amp;= x_n \sin 1 + y_n \cos 1.
\end{align*} $$

Now assume $(y_n)$ converges. Then since $\sin 1 \neq 0$, we have

$$ x_{n+1} = (y_{n+1} - y_n \cos 1) \cot 1 - y_n \sin 1$$

and hence $(x_n)$ also converges. Now let $(x_n, y_n) \to (\alpha, \beta)$. Then taking limit to the recursive formula we have

$$ \begin{align*}
\alpha &amp;= \alpha \cos 1 - \beta \sin 1 \\
\beta &amp;= \alpha \sin 1 + \beta \cos 1.
\end{align*} $$

Solving this system of linear equations give $(\alpha, \beta) = (0, 0)$. On the other hand, since

$$ x_n^2 + y_n^2 = 1, $$

we must have 

$$ \alpha^2 + \beta^2 = 1,$$

a contradiction! Therefore $(y_n)$ cannot converge. ////

Of course, we can say much more on $(y_n)$. For example, we can show that the set of limit points of $(y_n)$ is exactly $[-1, 1]$, and the Cesaro mean of $(y_n)$ is 0 from Weyl's criterion.



(This proof is from the book Problems in Real Analysis Advanced Calculus on the Real Axis.)
A known resultis that if {x_n} and {y_n} are two sequences such that y_n is divergent and x_n\y_n converges to 0 then x_n also diverges.
Take x_n=sin (n) and y_n=n then the result follows.