let be $ x \in \mathbb{R} $

I want to show that if $$ \lim_{n \rightarrow \infty } \sin(nx)=0 $$ then $$\sin(x)=0 $$

Do you have any ideas for ways to prove it? I don't know how to get there.. maybe using the fact that $\sin(a+b)= \sin(a)\cos(b)+\cos(a)\sin(b)..$ ?

thanks for any help!


$\sin (n+1)x=\sin (nx)\cos x+\cos (nx) \sin x$. So $\sin (nx) \to 0$ implies that $cos (nx) \sin x \to 0$. If $\sin x \neq 0$ this gives $ \cos (nx) \to 0$. But then $1=\sin^{2}(nx)+\cos^{2} (nx) \to 0$ , a contradiction.


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