How to solve this seemingly easy problem?

Question

In short, I need to prove that:

$\sin 2nx\not\to-\sin 2x\quad x\ne\frac{k\pi}2,k\in\Bbb Z,n\in\Bbb N\quad \text{as}\quad n\to\infty$

The biggest trouble is that I know little about $x$, not even knowing if it is a rational multiple of $\pi$ or not! This doesn't seem to be a hard problem, but I somehow get stuck.

Can you help me with that? I'd be very grateful!

PS: as little number theory involved as possible, please. I don't want to mess up with the irrationality of $\pi$.

Answer 1

First, in Demidovich's exercise book for mathematical analysis (problem 2553) there is a proof for $\sin (nx)\not\to 0, \forall x\not\in \pi\Bbb Z$ which goes as follows:

If $\lim\sin(nx)=0$ holds, then we also have $\lim\sin[(n+1)x]=0$. Note that $$\sin[(n+1)x]=\sin (nx)\cos x+\cos (nx)\sin x$$ and hence $$0=\lim\sin[(n+1)x]=\lim\{\sin (nx)\cos x+\cos (nx)\sin x\}$$ since $\lim\sin(nx)=0$, $\lim\sin(nx)\cos x=0$, which yiels $$0=\lim\cos (nx)\sin x=\lim\cos(nx)\quad(\sin x\ne0)$$ Therefore $$1=\lim[\sin^2(nx)+\cos^2(nx)]=0$$ which contradicts the hypothesis.

With the aid of this proof combined with Cauchy's criterion, I am now able to prove that $$\not\exists\lim\sin(nx)\in\Bbb R,\quad\forall x\not\in\pi\Bbb Z$$ the proof is as follows:

If $\exists\lim\sin(nx)\in\Bbb R$, according to Cauchy, $$\forall \epsilon>0,\exists N\in\Bbb N\quad \text{s.t.}\quad|\sin(mx)-\sin(nx)|<\epsilon,\quad\forall m>n>N$$ But $$|\sin(mx)-\sin(nx)|=\left|2\sin\left(\frac{(m-n)x}{2}\right)\cos\left(\frac{(m+n)x}{2}\right)\right|$$ For which now I set $m-n=p$ to be a fixed number, then $$|\sin(mx)-\sin(nx)|=\left|2\sin\left(\frac{px}{2}\right)\cos\left(nx+\frac p2x\right)\right|$$ since $x\not\in\pi\Bbb Z$, we have $$\left|2\sin\left(\frac{px}{2}\right)\right|=c\ne0$$ And, by following the tricks used by Demidovich, it is not hard to conclude that $$\cos\left(nx+\frac p2x\right)\not\to0$$ which means, however big $N$, I am always able to choose a $n>N$ such that $$\left|\cos\left(nx+\frac p2x\right)\right|>\epsilon$$ and hence $$\left|2\sin\left(\frac{px}{2}\right)\cos\left(nx+\frac p2x\right)\right|>c\epsilon$$ which indicates $\sin (nx)$ is not a Cauchy sequence, and hence diverges.

---

Tiny edit: to show $\cos(nx+px/2)\not\to0$, for a proof completely analogous to Demidovich's, $p$ should be specified to be $1$ to ensure neither $\sin px/2$ nor $\cos px/2$ vanishes for some $x\not\in\pi\Bbb Z$. ; )