Let $x_n\,(n\in\Bbb N)$ be the sequence defined by $$x_{n+1}=\frac{x_n}{\sqrt{x_n^2+1}+1}\tag 1$$ then it's well know that $2^nx_n\xrightarrow{n\to\infty}\arctan(x_0)$.
This gives a very simple recursive algorithm to manually compute $\arctan$ on a pocket calculator (like this) with $+,-,\times,\div,\sqrt{\phantom{x}}$ and one memory cell with keys M+, M-, MR.
In particular, to calculate the arcotangent of the number displayed on the screen of this calculator it's enough to press the following sequence of keys (see below for a proof): \begin{align} &\div\,=\,=\,\texttt{M+}\,\div\,=\,=\,\times\,=\\ &\texttt{While display > 0}\\ &\qquad+\,1\,=\,\sqrt{\phantom{x}}\,-\,1\,=\,\div\,2\,\times\,\texttt{MR}\,=\,\texttt{M+}\,\div\,\texttt{MR}\,=\\ &\texttt{MR}\,\div\,=\,= \end{align} In particular, we can compute $\pi$ with the following algorithm: \begin{align} &1\,\texttt{M+}\\ &\texttt{While display > 0}\\ &\qquad+\,1\,=\,\sqrt{\phantom{x}}\,-\,1\,=\,\div\,2\,\times\,\texttt{MR}\,=\,\texttt{M+}\,\div\,\texttt{MR}\,=\\ &\texttt{MR}\,\div\,4\,\div\,=\,= \end{align}
Note that at each ripetition of the while loop the same sequence of keys is pressed. A similar algorithm is possibile also for $\log,\arccos$ and the inverse hyperbolic cosine function.
Question. I'm ask if is possible to compute $\exp,\sin,\cos$ with an algorithm of the form \begin{align} &\texttt{S1}\\ &\texttt{While display > 0}\\ &\qquad\texttt{S2}\\ &\texttt{S3} \end{align} where $\texttt{S1},\texttt{S2}$ and $\texttt{S3}$ are finite sequences of keys among $$=,+,-,\times,\div,\sqrt{\phantom{x}},\texttt{M+},\texttt{M-},\texttt{MR},0,1,2,3,4,5,6,7,8,9$$
By completeness, I give a proof of the algorithm above. Assuming $x_0\neq 0$, let define $m_0=x_0^{-1}$ and \begin{align} m_{n+1} &=m_n\frac{\sqrt{x_n^2+1}+1}2\\ &=\frac 1{x_0}\prod_{k=0}^n\frac{\sqrt{x_k^2+1}+1}2 \end{align} so that \begin{align} 2^{n+1}x_{n+1} &=2^nx_n\left(\frac{\sqrt{x_n^2+1}+1}{2}\right)^{-1}\\ &=x_0\left(\prod_{k=0}^n\frac{\sqrt{x_k^2+1}+1}{2}\right)^{-1}\\ &=\frac 1{m_{n+1}} \end{align} hence $m_n^{-1}\xrightarrow{n\to\infty}\arctan(x_0)$. Let $d_n=x_n^2$ and $$d'_n=m_n\frac{\sqrt{d_n+1}-1}{2}$$ so that \begin{align} m_{n+1}&=m_n+d'_n\\ d_{n+1}&=\frac{d'_n}{m_{n+1}} \end{align}
Now assume that on the screen of the calculator we have the number $x_0$. By pressing $\div\,=\,=\,\texttt{M+}$ we compute $m_0=x_0^{-1}$ and we put it into the memory cell. Then $\div\,=\,=$ gives $x_0$ again and $\times\,=$ squares it giving $d_0=x_0^2$.
I claim that at each repetition of while loop we have stored $m_n$ in the memory cell and we display $d_n$ on the screen. This clearly holds of $n=0$. We argue by inductionon $n$; assume that $m_n$ is stored in the memory cell and $d_n$ is displayed on the screen.
The sequences of keys $+\,1\,=\,\sqrt{\phantom{x}}\,-\,1\,=\,\div\,2\,\times\,\texttt{MR}\,=$ gives $d'_n$ which is now displayed on the screen. Then by pressing $\texttt{M+}$ we add $d'_n$ to the memory cell $m_n$ giving $m_{n+1}=m_n+d'_n$ which is now stored in the memory cell. Finally $\,\div\,\texttt{MR}\,=$ computes $d_{n+1}$ which is now displayed on the screen, as claimed.
When we display $0$ on the screen we have $d_n\approx 0$, hence no better approximation can be computed and the while loop terminates.
The last keys $\texttt{MR}\,\div\,=\,=$ gives $m_{n+1}^{-1}\approx\arctan(x_0)$ as required.
