TITLE: Sine and Cosine Expansion Problem
QUESTION [8 upvotes]: We know:
$$\sin{x} = x - \frac{x^3}{3!} + \frac{x^3}{5!}-\dotsb$$
 and so on. Also,
$$\cos{x} = 1 - \frac{x^2}{2!} +\frac{x^4}{4!}-\dotsb$$
and so on.
With the help of these expansions we need to prove that $\sin^2 x+\cos^2 x=1$ .
I tried generalizing $\sin{x}$ as
$$\sum (-1)^n\frac{x^{2n+1}}{(2n+1)!}$$
and $\cos{x}$ as 
$$\sum (-1)^n\frac{x^{2n}}{(2n)!},$$
 then squaring and adding. But it didn't get through.
Please help!

REPLY [4 votes]: I think the easiest way to see it is to notice that $e^{ix}=\cos(x)+i\sin(x)$ which is easily verified by writing down the Taylor expansion of both sides. Then
$$\sin^2(x)+\cos^2(x)=e^{ix}e^{-ix}=1$$
If you want to verify $f(y)=e^{y}e^{-y}=1$, here's how: first show $\frac{d}{dx}e^y=e^y$ from the Taylor series. Then show $f'(y)=f(y)-f(y)=0$ by the chain rule. Conclude $f(y)$ is a constant, and plug in $y=0$ to get $f(y)=1$.