Approximating $\pi$ using the sine function If we have some approximation $x$ for $\pi$, it is possible to improve that approximation by calculating $\sin(x) + x$ if $x$ is sufficiently close to $\pi$. The reason why this works is that for $x \approx \pi$, $\sin(x) \approx \pi - x$ (note that $\sin'(\pi) = -1$), so $x + \sin(x) \approx x + \pi - x = \pi$.
I am interested in the number of good digits when approximating $\pi$ by iteratively applying this technique iteratively starting with the number $3$. In other words, I am interested in the following sequences:
$$
a_0=3; a_{n+1}=\sin(a_n)+a_n\\
b_n=\text{The number of digits of accuracy of }a_n
$$
The first few elements of $b$ are $\{0, 3, 10, 32, 99, 300, 902, 2702\}$. I did not find this sequence in OEIS. Interestingly, the number of correct digits seems to almost triple with every step.
Why does this method of approximating $\pi$ triple the number of accurate digits? If this approximation or sequence has been studied before, any pointers are welcome as well.
 A: 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 ($d_n=3d_{n-1}+0.778$).
A: 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.
A: 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 $a_n$.
A: Denote $c_n = a_n -  \pi$, then you ask how fast the sequence approaches zero. Now we have 
$$\begin{align}c_{n+1} &= \sin(\pi + c_n) + c_n\\ &= c_n - \sin(c_n)\\ &= \frac{c_n^3}{6} +o(c_n^3)\end{align}$$
So the decimal expansion of $c_{n+1}$ should have roughly three times more zeros than $c_n$, which explains the tripling of accurate digits.
A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\mbox{Newton-Raphson:}\
\left\{\begin{array}{rcl}
\ds{x_{0}} & \ds{=} & \ds{\color{#f00}{3}}
\\
\ds{x_{n}} & \ds{=} &
\ds{x_{n - 1} - {\sin\pars{x_{n - 1}} \over \cos\pars{x_{n - 1}}} =
x_{n - 1} - \tan\pars{x_{n - 1}}\,,\qquad n \geq 1}
\end{array}\right.}$
$\texttt{Mathematica}$:
(*Newton-Raphson*)
Clear[n, x];
Module[{n = 0, x = N[3,50]},
While[n++ < 20, x -= N[Tan[x], 50]];
N[x, 50]]

Result:3.1415926535897932384626433832795028841971693993751
All the digits are 'correct'.
A: Your method is an improvement over Newton's method, which would be to look at the sequence $u_n$ defined by $u_0=3$ and
$$u_{n+1}=u_n-\dfrac{\sin(u_n)}{\sin'(u_n)}$$
You're using some apriori knowledge about the root, hence getting a quicker convergence. Newton's method is known for doubling the number of accurate digits in each iteration.
A: This iterative approach for $\pi$ was considered by Daniel Shanks in a 1-page note:
"Improving an approximation for pi." Amer. Math. Monthly 99 (1992), no. 3, 263. He does not spell out the cubic convergence. 
