<s>[INST] 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?
 [/INST] Note that $\sin$ is the unique solution of $y'' = -y$ subject to $y(0) = 0, y'(0) = 1$. Note that $x \mapsto \sin (x+2 \pi)$ also satisfies the same
differential equation, so the question boils down to showing that
$\sin (2 \pi) = 0, \cos(2 \pi) = 1$ (where $\cos = \sin'$).
Consider $\eta(x) = \sin^2 x + \cos^2 x$, we see $\eta(0) = 1$ and $\eta(x)'= 0$, hence $\sin^2 x + \cos^2 x =1$ for all $x$, and so we only need to establish that $\cos (2 \pi) = 1$.
It is straightforward to establish that there is a smallest $T>0$ such that $
\cos T = 0$. Rudin has a nice method in "Real & Complex Analysis". Alternatively, suppose $\cos x >0$ for all $x \ge 0$. We  have
$\cos 0 = 1$, and so we have some $x_0>0$ such that $\sin x_0 >0$. By
presumption, $\sin $ is strictly increasing on $x \ge 0$, and since $\cos' x = - \sin x$, we have $\cos' x \le - \sin x_0$ for all $x \ge x_0$,
which contradicts $\cos x >0$ for all $x \ge 0$. Since $\cos$ is continuous,
there is a smallest positive number $T$ such that $\cos T = 0$, and we can see that $\sin T = 1$, since $\sin$ is increasing on $[0,T)$. 
We define $\pi = 2 T$.
Let $\zeta(x) = \cos ( {\pi \over 2}-x)$. We note that $\zeta(0) = 0, \zeta'(0) = 1$ and
$\zeta'' = - \zeta$, and so by uniqueness we have $\sin x= \cos ({\pi \over 2}-x) = \cos (x-{\pi \over 2})$ for all $x$ (since $\cos $ is even), and
differentiating gives
$\cos x = -\sin (x-{\pi \over 2})$ (since $\cos' = -\sin$). It is
straightforward to establish that $\cos (2 \pi) = 1$ from this.
Addendum: To show that the solution of $y''+y = 0$ is unique we could use the Picard Lindelöf theorem. Alternatively, suppose $y_1,y_2$ are two solutions with
the same initial conditions and let $\delta = y_1-y_2$. Let $V = \delta^2 + (\delta')^2$ and note that $V' = 0$. Since $V(0) = 0$, we have $V(x) = 0$ for all $x$, that is, $y_1 = y_2$.
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