The behavior of $\sin(\sin(\sin(\cdots)))$, $\cos(\cos(\cos(\cdots)))$, and $\sin(\cos(\sin(\cos(\cdots))))$ I was playing around with desmos, when i saw this

I think that:

*

*$\sin(\sin(\sin(\cdots)))$ will approach zero veeeery slowly, also approaching a square wave.


*$\cos(\cos(\cos(\cdots)))$ will approach $0.7389$.


*$\sin(\cos(\sin(\cos(\cdots))))$ will approach $0.948$.
Are these hypothesis correct? Is there any formal proof?
 A: By the fixed-point theorem, when an iteration like
$$x_{n+1}=f(x_n)$$ converges, it converges to a solution of $f(x)=x$.
Here is a plot of your $f$'s, which confirm the values you found:

A convergence criterion is $|f'(x)|<1$, which is achieved by your last two functions. Convergence is linear (the number of exact digits grows proportionally to the $n$).
The case of the sine is a little more difficult. But as $|\sin(x)|<|x|$, convergence to $0$ is guaranteed.
If we assume $\sin^{(n)}(x)\sim cn^\alpha$, we have
$$\sin^{(n+1)}(x)=\sin^{(n)}(x)-\frac{(\sin^{(n)}(x))^3}6+o((\sin^{(n)}(x))^3)$$
corresponding to
$$c(n+1)^\alpha\sim cn^\alpha-\frac c6n^{3\alpha},$$ which gives $\alpha=-\frac12$ by identifying the second terms, and
$$\sin^{(n)}(x)\sim\frac c{\sqrt n}.$$
This convergence is extremely slow.

As you probably already figured out, the iterations of $\cos(\sin(\cos(\sin(\cdots))))$ will converge to the value of $\cos(x)$ where $x$ is the limit of $\sin(\cos(\sin(\cos(\cdots))))$.
A: For the iterated sine you can use the following: Since $|\sin(x)|\leqslant 1$ you can deduce that $\sin(\sin(x))\in[-\sin(1),\sin(1)]$ and you can easily see that $\sin(1)<1$ using Taylor expansion. Now if you take a sequence given by $\varepsilon_n = \sin(\varepsilon_{n-1})$ with say $0<\varepsilon_0<1$, then you get (since for all $0<x<1$ you have $0<\sin(x)<x$) a decreasing sequence $\varepsilon_n$ which is bounded below by $0$. Then you know that your sequence converges: In the limit you will have $\sin(\varepsilon_\infty)=\varepsilon_\infty$ and hence $\varepsilon_\infty=0$.
