We can see intuitively that $ f(x)=\sin\left(\frac{\pi}{2}\sin\left(\frac{\pi}{2}\sin\left(\frac{\pi}{2}\cdots\sin{x}\cdots\right)\right)\right) $ is the square wave with period $2\pi$ and has the value $0$ at the jumps, i.e $ f(x)= \begin{cases} 0 & \text{if } \frac{x}{\pi}\in\mathbb{Z}\\ \mathrm{sign}(\sin{x}) & \text{otherwise} \end{cases} $ Look at this graph of $x$ and $\sin{\frac{\pi}{2}x}$ to see why :
But $f(x)$ is then also exactly equal to the Fourier series of the square wave with period $2\pi$ since Dirichlet conditions assure that the series converges to $0$ (the midpoint) at the jumps as do $f(x)$.
Hence we might be able to show that, $ \sin\left(\frac{\pi}{2}\sin\left(\frac{\pi}{2}\sin\left(\frac{\pi}{2}\cdots\sin{x}\cdots\right)\right)\right)=\frac{4}{\pi}\sum_{k=0}^{\infty}\frac{\sin{(2k+1)x}}{2k+1} $
Does anyone have an idea of how to prove this directly? Are there other Fourier series that are equal to a recursive formula of trigonometric functions?
Restated, the problem is to show that if $ f_0(x)=\sin{x},\quad\text{and}\quad f_n(x)=\sin{\left(\frac{\pi}{2}f_{n-1}(x)\right)} $ then, $ \lim_{n\to\infty}{f_n(x)}=\frac{4}{\pi}\sum_{k=0}^{\infty}\frac{\sin{(2k+1)x}}{2k+1} $