I'll write it formally:
Let $f(1,x) = \sin(x)$,
Let $f(n+1,x) = \sin\bigl(f(n,x)\bigr)$ for $n\in \Bbb N$ with $n>1$.
  What is the limit as $n \to \infty$?

It's fairly easy to prove that the absolute value of $f(n,x)$ is decreasing for every constant $x$ as $n$ is increasing, so the limit exists. From what I calculated, it seems like $f(\infty,x)= 0$ for all $x$. Does someone have a proof for that?

Follow up: The limit $f(n,x)\rightarrow 0$ may be a bit trivial. What if we rescale $x$ and ask for the limit of $n^\alpha f(n,n^{-\alpha}x)$ as $n\rightarrow\infty$? For which $\alpha$ does the limit exist and what is it?

I'll write it formally: Let $\sin^{\circ\, 1}(x) = \sin(x)$ and $\sin^{\circ n+1}(x) = \sin\bigl(\sin^{\circ n}(x)\bigr)$ for $n\in \Bbb N$ with $n>1$. What is the limit as $n \to \infty$?

It's fairly easy to prove that the absolute value of $f(n,x)$ is decreasing for every constant $x$ as $n$ is increasing, so the limit exists. From what I calculated, it seems like $f(\infty,x)= 0$ for all $x$. Does someone have a proof for that?

Follow up: The limit $\sin^{\circ n}(x)\rightarrow 0$ may be a bit trivial. What if we rescale $x$ and ask for the limit of $$n^\alpha \sin^{\circ n}(n^{-\alpha}x)$$ as $n\rightarrow\infty$? For which $\alpha$ does the limit exist and what is it?