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.
 A: Nice proof! To address your question,

Do Taylor's series of those functions depend on this identity? Because if they do, the proof will be circular.

You are right to be worried about circularity! However, which concepts or theorems depend on which others is a matter of certain flexibility. Often, we take one thing to be the definition of a concept, and then have to prove the other properties as theorems -- but we could alternately have used some other property as a definition, and then the original definition would have to be a theorem.
Specific to your case, I have seen definitions of $\sin$ and $\cos$ where we start by defining $\sin$ using arclength, then we define $\cos x$ to satisfy $\cos^2 x + \sin^2 x = 1$. If we take this approach, certainly, there is nothing to prove. However, this is not the only possible approach! It is also common to define $\sin$ and $\cos$ using their Taylor series. Under this approach, you have given a very nice proof that $\sin^2 x + \cos^2 x = 1$.
Your proof could also be valid if we define $\sin$ and $\cos$ to be a basis of functions satisfying $f''(x) = -f(x)$.
In summary, it depends on what you define $\sin$ and $\cos$ to be; however, your proof is not necessarily circular. And it is a nice example of deriving a result about some functions from their Taylor series.
A: No, they do not depend on the Pythagoren identity. Using the differential equations that define $\sin$ and $\cos$


*

*$\cos(0) = 1$

*$\sin(0) = 0$

*$\cos' = -\sin$

*$\sin' = \cos$
it's very easy to show that $\forall x \in \mathbb{R}\cos(x) = \sum_{i = 0}^\infty \frac{(-1)^ix^{2i}}{(2x)!}$ and $\forall x \in \mathbb{R}\sin(x) = \sum_{i = 0}^\infty \frac{(-1)^ix^{2i + 1}}{(2x + 1)!}$.
A: 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:
$$ \sin^2(\rho)+\cos^2(\rho) = 1.\tag{4}$$
