how can we convert sin function into continued fraction?

how can we convert sin function into continued fraction ?

for example

http://mathworld.wolfram.com/EulersContinuedFraction.html

how can we convert sin to simmilar continued fraction ??

and what about sinh and cosh ? arcsin ? arctan ? cos ? arccos ??

in general , how can convert any function to continued fraction ???

my friend asked me this question , so i hope that you help me to be enabled to help him

thanx for all of you

Solutions Collecting From Web of "how can we convert sin function into continued fraction?"

We will proceed as in this answer.

Define
$$
P_n(x)=\sum_{k=0}^\infty\frac{4^{k+n}-\sum\limits_{j=1}^n\binom{2k+2n+1}{2j-1}}{(2k+2n+1)!}(-x^2)^k\tag{1}
$$
Then
$$
\begin{align}
\frac{P_{n-1}(x)}{P_n(x)}
&=\frac
{\displaystyle\sum_{k=0}^\infty\frac{4^{k+n-1}-\sum\limits_{j=1}^{n-1}\binom{2k+2n-1}{2j-1}}{(2k+2n-1)!}(-x^2)^k}
{\displaystyle\sum_{k=0}^\infty\frac{4^{k+n}-\sum\limits_{j=1}^n\binom{2k+2n+1}{2j-1}}{(2k+2n+1)!}(-x^2)^k}\\[12pt]
&=\color{#C00000}{-x^2+}\frac
{\displaystyle\sum_{k=0}^\infty\frac{\color{#C00000}{\binom{2k+2n-1}{2n-1}}}{(2k+2n-1)!}(-x^2)^k}
{\displaystyle\sum_{k=0}^\infty\frac{4^{k+n}-\sum\limits_{j=1}^n\binom{2k+2n+1}{2j-1}}{(2k+2n+1)!}(-x^2)^k}\\[12pt]
&=-x^2+\frac
{\displaystyle\sum_{k=0}^\infty\color{#C00000}{\frac{2n(2n+1)\binom{2k+2n+1}{2n+1}}{(2k+2n+1)!}}(-x^2)^k}
{\displaystyle\sum_{k=0}^\infty\frac{4^{k+n}-\sum\limits_{j=1}^n\binom{2k+2n+1}{2j-1}}{(2k+2n+1)!}(-x^2)^k}\\[12pt]
&=\color{#C00000}{2n(2n+1)}-x^2\color{#C00000}{-}\frac
{\displaystyle\sum_{k=0}^\infty\frac{2n(2n+1)\color{#C00000}{\left[4^{k+n}-\sum\limits_{j=1}^{n+1}\binom{2k+2n+1}{2j-1}\right]}}{(2k+2n+1)!}(-x^2)^k}
{\displaystyle\sum_{k=0}^\infty\frac{4^{k+n}-\sum\limits_{j=1}^n\binom{2k+2n+1}{2j-1}}{(2k+2n+1)!}(-x^2)^k}\\[12pt]
&=2n(2n+1)-x^2\color{#C00000}{+2n(2n+1)x^2}\frac
{\displaystyle\sum_{k=0}^\infty\color{#C00000}{\frac{4^{k+n+1}-\sum\limits_{j=1}^{n+1}\binom{2k+2n+3}{2j-1}}{(2k+2n+3)!}}(-x^2)^k}
{\displaystyle\sum_{k=0}^\infty\frac{4^{k+n}-\sum\limits_{j=1}^n\binom{2k+2n+1}{2j-1}}{(2k+2n+1)!}(-x^2)^k}\\[12pt]
&=2n(2n+1)-x^2+2n(2n+1)x^2\color{#C00000}{\left/\frac{P_n(x)}{P_{n+1}(x)}\right.}\tag{2}
\end{align}
$$
As I suggested in chat, consider
$$
\begin{align}
\sin(x)
&=\frac{\sin(2x)}{2\cos(x)}\\
&=\frac
{\displaystyle x\sum_{k=0}^\infty\frac{4^k(-x^2)^k}{(2k+1)!}}
{\displaystyle\sum_{k=0}^\infty\frac{(-x^2)^k}{(2k)!}}\\
&=x\left/\left(\frac
{\displaystyle\sum_{k=0}^\infty\frac{(-x^2)^k}{(2k)!}}
{\displaystyle \sum_{k=0}^\infty\frac{4^k(-x^2)^k}{(2k+1)!}}
\right)\right.\\
&=x\left/\left(1+x^2\left/\frac{P_0(x)}{P_1(x)}\right.\right)\right.\tag{3}
\end{align}
$$
$(2)$ and $(3)$ lead us to the continued fraction
$$
\sin(x)=\cfrac{x}{1+\cfrac{x^2}{2\cdot3-x^3+\cfrac{2\cdot3x^2}{\ddots\lower{6pt}{2n(2n+1)-x^2+\cfrac{2n(2n+1)x^2}{P_n(x)/P_{n+1}(x)}}}}}\tag{4}
$$

I learned how to do this from this document (look for Theorem I)
https://people.math.osu.edu/sinnott.1/ReadingClassics/continuedfractions.pdf

Interestingly, this looks to be a translation of Euler’s original work. I find him easy to understand, he avoids confusing his message in clumsy notation.