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?

Using term-wise derivation, we define the auxiliary sum $$e^{ix}=\sin'(x)+i\sin(x)=\sum_{k=0}^\infty\frac{(ix)^k}{k!}.$$
By a straightforward manipulation of this sum using the Binomial theorem,
$$e^{i(x+t)}=\sum_{k=0}^\infty\frac{(i(x+t))^k}{k!}=\sum_{k=0}^\infty\sum_{l=0}^k\frac{k!i^kx^lt^{k-l}}{k!l!(k-l)!}=\sum_{l=0}^\infty\sum_{j=0}^\infty\frac{(ix)^l(it)^{j}}{l!j!}=e^{ix}e^{it}.$$
Then we evaluate $$\sin(1)=1-\frac1{3!}+\text{ positive pairs of terms }\cdots>0$$
$$\sin(4)=4-\frac{4^3}{3!}+\frac{4^5}{5!}-\frac{4^7}{7!}+\frac{4^9}{9!}-\text{ positive pairs of terms }\cdots<0,$$
to show that $\sin$ has a root in $(1,4)$, let $\pi$.
As $|e^{i\pi}|^2=e^{i\pi}\left(e^{i\pi}\right)^*=e^{i\pi}e^{-i\pi}=e^0=1$, this root is such that
$$e^{i\pi}=\pm1,$$ and to avoid the possible negative, we use $$e^{i2\pi}=\left(e^{i\pi}\right)^2=1.$$Thus there is a real $\pi\in(1,4)$ such that
$$\sin(x+2\pi)=\Im\left(e^{ix+i2\pi}\right)=\Im\left(e^{ix}\right)=\sin(x).$$
