Approximating $\pi$ using the sine function

The Taylor Series for $\sin(x)$ for $x$ near $\pi$ says $$ \sin(x)=\sin(\pi-x)=(\pi-x)-\frac{(\pi-x)^3}6+O\!\left((\pi-x)^5\right) $$ Thus $$ x+\sin(x)-\pi=\frac{(x-\pi)^3}6+O\!\left((\pi-x)^5\right) $$ That is, $$ x_{n+1}-\pi\sim\frac{(x_n-\pi)^3}6 $$ which means the number of correct digits more than triples with each iteration ($d_n=3d_{n-1}+0.778$).

The Taylor expansion at $x=\pi$ is $$\sin(x)= \pi-x + \frac{1}{6}(x-\pi)^3- O((x-\pi)^4)$$ $$\sin(x) +x = \pi + \frac{1}{6}(x-\pi)^3- O((x-\pi)^4)$$ Therefore $$a_{n+1}-\pi = \sin(a_n)+a_n-\pi = \frac{1}{6}(a_n-\pi)^3- O((a_n-\pi)^4)$$

This means that the correct digits triple with each step, after convergence has set-it.

Note that $$|a_{n+1}-\pi|=|a_{n}+\sin(a_n)-\pi|leq =|(a_{n}-\pi)-\sin(a_n-\pi)|\leq C\cdot\frac{|a_{n}-\pi|^3}{6}$$ because $\sin(t)=t-\frac{t^3}{6}+O(t^5)$ as $t\to 0$. So the order of convergence is $3$ So the decimal expansion of $a_{n+1}$ should have about three times more zeros than $a_n$.