If $g(x)$ is a composition of arbitrary finite sequence of $\sin(x)$ and $\cos(x)$, does $x_n = g(x_{n-1})$ converge to unique $x$?

Question

Suppose we have a sequence of $k$ functions $g_i$ for $1 \leq i \leq k$ where each $g_i(x)$ is either $\sin x$ or $\cos x$. Define $g(x) = g_1(g_2(\ldots(g_k(x)))$ to be the composition of the $g_i$. If $x_0$ is arbitrary, does $x_n = g(x_{n-1})$ converge to a unique point independent of $x_0$?

Answer 1

Yes. Note that the image of $g$ lies in $[-1, 1]$ so we can wlog assume that $x_0 \in [-1, 1]$. On this interval the inequalities $|\sin(x) - \sin(y)| < |x - y|$ and $|\cos(x) - \cos(y)| < |x - y|$ hold for all $x \ne y$. The assertion now follows from this version of the Banach fixed point theorem.