How can you derive $\sin(x) = \sin(x+2\pi)$ from the Taylor series for $\sin(x)$? \begin{eqnarray*}
\sin(x) & = & x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots\\
\sin(x+2π) & = & x + 2\pi - \frac{(x+2π)^3}{3!} + \frac{(x+2π)^5}{5!} - \ldots \\
\end{eqnarray*}
Those two series must be equal, but how can you show that by only manipulating the series?

Define $f(x)$ as the residual:
$$\sin(x+2\pi) = x + 2\pi - \frac{(x+2\pi)^3}{3!} + \frac{(x+2\pi)^5}{5!}+\ldots$$
$$\equiv x - \frac{x^3}{3!} + \frac{x^5}{5!}+\ldots + f(x)=\sin(x)+f(x)$$
Now prove it is zero:
$$f(x) = 2\pi -\frac{1}{3!}\left((2\pi)^3 + 3(2\pi)^2 x + 3(2\pi)x^2 \right)+
\frac{1}{5!}\left( (2\pi)^5 +5x (2\pi)^4+10x^2(2\pi)^3+10x^3(2\pi)^2+5(2\pi)x^4\right)+\ldots$$
Extract all powers of $x$ lower than $2$, and define yet another residual, $g(x)$:
$$f(x)=\sin(2\pi) + x\sum_{k=1}^\infty \frac{(-1)^k(2\pi)^{2k}}{(2k)!}+g(x)$$
$$=\sin(2\pi) +x(1-\cos(2\pi))+g(x)$$
The first terms are zero, and now we must prove that the remaining residual $g(x)$ is also zero, ad infinitum. 
A more general continuation to follow... 
