Do the Taylor series of $\sin x$ and $\cos x$ depend on the identity $\sin^2 x + \cos^2 x =1$? I had this crazy idea trying to prove the Pythagorean trigonometric identity;$$\sin^2x+\cos^2x=1$$by squaring the infinite Taylor series of $\sin x$ and $\cos x$. But it came out quite beautiful, involving also a combinatorics identitie. 
The proof:
$$\sin x=\frac{x}{1}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...=\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}\\\\\sin^2x=x^2-x^4\left (\frac{1}{1!3!}+\frac{1}{3!1!}\right )+x^6\left (\frac{1}{1!5!}+\frac{1}{3!3!}+\frac{1}{5!1!}\right )-...\\\\\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}\\\\\cos^2x\!=\!1\!-\!x^2\left(\!\frac{1}{0!2!}\!+\!\frac{1}{2!0!}\!\right)\!+\!x^4\left(\!\frac{1}{0!4!}\!+\!\frac{1}{2!2!}\!+\!\frac{1}{4!0!}\!\right)\!-\!x^6\left(\!\frac{1}{0!6!}\!+\!\frac{1}{2!4!}\!+\!\frac{1}{4!2!}\!+\!\frac{1}{6!0!}\!\right)\!+...$$We should have shown that the series for both $\sin x$ and $\cos x$ converge absolutely (since we changed the arrangement), but it's obvious since the absolute value of all terms of $\sin x+\cos x$ add up to $e^x$.$$\sin^2x+\cos^2x=\\=1-x^2\left(\frac{1}{0!2!}-\frac{1}{1!1!}+\frac{1}{2!0!}\right)+x^4\left(\frac{1}{0!4!}-\frac{1}{1!3!}+\frac{1}{2!2!}-\frac{1}{3!1!}+\frac{1}{4!0!}\right)-x^6\left(\frac{1}{0!6!}-\frac{1}{1!5!}+\frac{1}{2!4!}-\frac{1}{3!3!}+\frac{1}{4!2!}-\frac{1}{5!1!}+\frac{1}{6!0!}\right)+...=\\\\=1+\sum_{n=1}^{\infty}(-1)^nx^{2n}\sum_{k=0}^{2n}\frac{(-1)^k\binom{2n}{k}}{(2n)!}$$
Since we can show easily that $\sum_{i=0}^n(-1)^i\binom{n}{i}=0$ by expanding $(1-1)^n$ using Binom's formula. So:$$\sin^2x+\cos^2x=1-0+0-0+...=1$$
I think it's beautiful. I just wanted to ask, do Taylor's series of those functions depend on this identity? Because if they do, the proof will be circular.

You are just showing that the Pythagorean theorem is a consequence of some property of the (complex) exponential function, there is no circularity in such argument. For instance, we may define, for any $z\in\mathbb{C}$,
$$ f(z)=e^{z}=\sum_{n\geq 0}\frac{z^n}{n!} \tag{1}$$
and prove through a combinatorial argument that such a function fulfills $e^{z}\cdot e^{w}=e^{z+w}$.
Since the Taylor coefficients at $0$ of such analytic function are real, we have $f(\bar{z})=\overline{f(z)}$, hence for any $\rho\in\mathbb{R}$
$$ \left\| e^{i\rho}\right\| = e^{i\rho}\cdot e^{-i\rho} = e^0 = 1. \tag{2}$$
If we define $\cos(\rho)$ and $\sin(\rho)$ as the real/imaginary part of $e^{i\rho}$, we get that
$$ \sin(\rho)=\sum_{n\geq 0}\frac{(-1)^n \rho^{2n+1}}{(2n+1)!},\qquad \cos(\rho)=\sum_{n\geq 0}\frac{(-1)^n \rho^{2n}}{(2n)!}\tag{3} $$
and $(2)$ can be read as:
$$ \sin^2(\rho)+\cos^2(\rho) = 1.\tag{4}$$
