Prove that the iteration of $\sin(x)$ goes to zero as $n$ goes to $\infty$

Question

Basically let $S(x)=\sin(x)$ such that $S^2(x)=\sin(\sin(x))$ and $S^3(x)=\sin(\sin(\sin(x)))$ and so on until $S^n(x)=\sin(\sin(\ldots\sin(x)\ldots))$

Prove that $S^n(x)\rightarrow 0$ as $n\rightarrow\infty$

Answer 1

As you stated, we have $S(x)=\sin x$ and $S^{n+1}(x)=\sin(S^n(x))$. The sequence $\{|S^n(x)|\}$ is bounded and we have $|\sin(x)|\leq|x|$ for all $x$, then $|S^{n+1}(x)|=|\sin(S^n(x))|\leq |S^n(x)|$: the sequence is decreasing. Hence the limit $\lim_{n\to\infty}|S^n(x)|$ exist, call it $L$. Then

$$\lim_{n\to\infty}|S^{n+1}(x)|=\lim_{n\to\infty}|\sin(S^n(x))|$$

$$L=|\sin(L)|$$

Then $L=0$, the only fixed point of the function $|\sin(x)|$.