<s>[INST] 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.
 [/INST] 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:
$$ \sin^2(\rho)+\cos^2(\rho) = 1.\tag{4}$$
</s>