How to give a rigorous proof to the divergence of $\sin nx$

Question

It's easy to know that sequence $\{\sin (nx)\} \ (x\neq k \pi)$ does not converge , but when I write the proof, I find it hard to give a rigorous one. Could someone please help me out here? A hint may help as well. Thanks in advance.

Answer 1

Claim: If $\sin (nx)$ converges as $n\to\infty$, then $\sin x=0$.

Proof: Use the sum and difference formulas $\sin(A+B)=\sin A\cos B + \cos A\sin B$ and $\sin(A-B)=\sin A\cos B-\cos A\sin B$ to write $$ \sin ((n+1)x) = \sin (nx)\cos x + \cos (nx)\sin x\tag1 $$ and $$ \sin( (n-1)x) = \sin (nx)\cos x - \cos (nx)\sin x.\tag2 $$ If $\sin (nx)$ converges to a limit $L$ as $n\to\infty$, then $\sin((n+1)x)$ and $\sin((n-1)x)$ converge to the same $L$. Add formulas (1) and (2), let $n\to\infty$, and deduce that $$2L=2L\cos x.$$

It follows that $\cos x=1$, or $L=0$. The case $\cos x=1$ immediately leads to $\sin x=0$.

So suppose $L=0$. Subtract (2) from (1), square both sides of the result, and take the limit as $n\to\infty$. Conclude that $\cos^2(nx)\cdot\sin^2 x$ converges to $0$ as $n\to\infty$. Since $\cos^2(nx)=1-\sin^2(nx)$ tends to $1$, conclude $\sin^2 x=0$ and therefore $\sin x=0$.

Answer 2

Here is a slightly different proof:

Let's assume that $f_n(x):=\sin(nx)$ converges for a $x\neq m\pi, m\in\mathbb{Z}$. Then $\sin^2(nx)$ and $\cos^2(nx)$ also converge. Using the formula $\cos(2x)=\cos^2(x)-\sin^2(x)$ proves that $\cos(nx)$ converges. Now we see that \begin{align*} &\lim\limits_{n\to\infty}\cos(x)=\lim\limits_{n\to\infty}\cos((n+1)x-nx)\\ &=\lim\limits_{n\to\infty}\cos((n+1)x)\cos(nx)+\lim\limits_{n\to\infty}\sin((n+1)x\sin(nx)\\ &=\lim\limits_{n\to\infty}\cos^2(nx)+\lim\limits_{n\to\infty}\sin^2(nx)=1. \end{align*} This is a contradiction, so $f_n(x)$ can't converge if $x\neq m\pi, m\in\mathbb{Z}$.