I am aware of Euler's continued fraction:
$$a_0+a_0a_1 + a_0a_1a_2 + a_0a_1a_2a_3 +\cdots = \cfrac{a_0}{1-\cfrac{a_1}{1+a_1-\cfrac{a_2}{1+a_2-\cfrac{a_3}{1+a_3-\ddots}}}}$$ (inductive proof)
I am also aware of the Taylor series representations of $\sin x$ and $\cos x$.
$$\begin{align}\sin x &= \sum_{i=0}^\infty \cfrac{(-1)^ix^{2i+1}}{(2i+1)!} = x - \cfrac{x^3}{3!}+\cfrac{x^5}{5!}-\cfrac{x^7}{7!}+\cdots\\ \cos x &= \sum_{i=0}^\infty \cfrac{(-1)^ix^{2i}}{(2i)!}=1-\cfrac{x^2}{2!}+\cfrac{x^4}{4!}-\cfrac{x^6}{6!}+\cdots\end{align}$$
So that's when I thought: what would happen if I combine these together?
For $\sin x$, I let $a_0=x$, $a_1=-\cfrac{x^2}{2\cdot 3}$, $a_2=\cfrac{x^2}{4\cdot 5}$, $a_3=-\cfrac{x^2}{6\cdot 7}$, thus getting
$$\begin{align}\sin x &=\cfrac{x}{1+\cfrac{x^2\div (2\cdot 3)}{1+x^2\div (2\cdot 3)-\cfrac{x^2\div (4\cdot 5)}{1+x^2\div (4\cdot 5)+\cfrac{x^2\div (6\cdot 7)}{1+x^2\div (6\cdot 7)-\ddots}}}} \\ &=\cfrac{x}{1+\cfrac{x^2}{x^2+(2\cdot 3)\Bigg\{1-\cfrac{x^2}{x^2+(4\cdot 5)\left\{1+\cfrac{x^2}{x^2+(6\cdot 7)\big\{1-\ddots\big\}}\right\}}\Bigg\}}}\end{align}$$
We could similarly derive a continued fraction for $\cos x$. However, I noticed a way to write $\sin 1$ and $\cos 1$ as continued fractions, with a much nicer form: $$\begin{align}\sin 1 &= \cfrac{1}{1+\cfrac{1!^2}{3!-1!+\cfrac{3!^2}{5!-3!+\cfrac{5!^2}{7!-5!+\cfrac{7!^2}{9!-7!+\ddots}}}}} \\ \cos 1 &= \cfrac{1}{1+\cfrac{0!^2}{2!-0!+\cfrac{2!^2}{4!-2!+\cfrac{4!^2}{6!-4!+\cfrac{6!^2}{8!-6!+\ddots}}}}}\end{align}$$ Methinks these could be derived by a combination of Euler's continued fraction and the Taylor series, but this looks different and way nicer than my results when I plug in $x=1$. How would one be able to derive the two latter continued fractions in order to prove them (if true)?
Thanks.