Prove the divergence of the sequence $\left\{ \sin(n) \right\}_{n=1}^{\infty}$.

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

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

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

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

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.

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(This proof is from the book Problems in Real Analysis Advanced Calculus on the Real Axis.)

If $\alpha$ is an irrational number, the numbers $n\alpha$, considered mod $1$, are dense in $[0,1]$. (A stronger result is that they are equidistributed.) Thus the numbers $2n/\pi$ are dense in $[0,1]$, and therefore the integers are dense mod $\pi/2$. It follows that $\sin n$ diverges.

Imagine marching around the circumference of the unit circle in steps of arc length $1$. Then $\sin(n)$ is the $y$ coordinate at the $n$th step. Since $\pi\gt1$, the $y$ coordinate will be positive infinitely often and negative infinitely often. Consequently the limit of $\sin(n)$, if it exists, would have to be $0$. But for the same reason ($\pi\gt1$), $\sin(n)\ge\sin({\pi-1\over2})=\sin(\pi-{\pi-1\over2})$ for infinitely many $n$, and hence the limit, if it exists, would have to be at least $\sin({\pi-1\over2})$, which is greater than $0$. These contradictory requirements show that the limit does not exist.

Remark: This proof does not rely on $\pi$ being an irrational number; nor does it use any trig identities beyond the simplest symmetries of the sine function.