<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] With series manipulation, we can get
$$
\begin{align}
\sin(x+y)
&=\sum_{k=0}^\infty(-1)^k\frac{(x+y)^{2k+1}}{(2k+1)!}\tag{1a}\\
&=\sum_{k=0}^\infty\sum_{j=0}^{2k+1}\frac{(-1)^k}{(2k+1)!}\binom{2k+1}{j}x^jy^{2k+1-j}\tag{1b}\\
&=\sum_{k=0}^\infty\sum_{j=0}^{2k+1}(-1)^k\frac{x^j}{j!}\frac{y^{2k+1-j}}{(2k+1-j)!}\tag{1c}\\
&=\sum_{k=0}^\infty\sum_{j=0}^k(-1)^j\frac{x^{2j+1}}{(2j+1)!}(-1)^{k-j}\frac{y^{2k-2j}}{(2k-2j)!}\tag{1d}\\
&+\sum_{k=0}^\infty\sum_{j=0}^k(-1)^j\frac{x^{2j}}{(2j)!}(-1)^{k-j}\frac{y^{2k-2j+1}}{(2k-2j+1)!}\tag{1e}\\
&=\sum_{j=0}^\infty(-1)^j\frac{x^{2j+1}}{(2j+1)!}\sum_{k=0}^\infty(-1)^{k}\frac{y^{2k}}{(2k)!}\tag{1f}\\
&+\sum_{j=0}^\infty(-1)^j\frac{x^{2j}}{(2j)!}\sum_{k=0}^\infty(-1)^k\frac{y^{2k+1}}{(2k+1)!}\tag{1g}\\[6pt]
&=\sin(x)\cos(y)+\cos(x)\sin(y)\tag{1h}
\end{align}
$$
Explanation:
$\text{(1a)}$: series definition
$\text{(1b)}$: binomial theorem
$\text{(1c)}$: break binomial coefficient into factorials
$\text{(1d)}$: add the odd powers of $x$ from $\text{(1c)}$
$\text{(1e)}$: to the even powers of $x$ from $\text{(1c)}$
$\text{(1f)}$: change the order of summation and substitute $k\mapsto k+j$
$\text{(1g)}$: change the order of summation and substitute $k\mapsto k+j$
$\text{(1g)}$: series definitions

We can prove that $\cos(2\pi)=1$ and $\sin(2\pi)=0$ using a couple of derivatives and some geometry. First,
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\sin(x)
&=\frac{\mathrm{d}}{\mathrm{d}x}\sum_{k=0}^\infty(-1)^k\frac{x^{2k+1}}{(2k+1)!}\\
&=\sum_{k=0}^\infty(-1)^k\frac{x^{2k}}{(2k)!}\\[6pt]
&=\cos(x)\tag{2}
\end{align}
$$
and
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\cos(x)
&=\frac{\mathrm{d}}{\mathrm{d}x}\left(1+\sum_{k=1}^\infty(-1)^k\frac{x^{2k}}{(2k)!}\right)\\
&=\sum_{k=1}^\infty(-1)^k\frac{x^{2k-1}}{(2k-1)!}\\
&=\sum_{k=0}^\infty(-1)^{k+1}\frac{x^{2k+1}}{(2k+1)!}\\[6pt]
&=-\sin(x)\tag{3}
\end{align}
$$
Using $(2)$ and $(3)$, we get
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\left(\sin^2(x)+\cos^2(x)\right)
&=2\sin(x)\cos(x)-2\cos(x)\sin(x)\\
&=0\tag{4}
\end{align}
$$
Equation $(4)$ says that $\sin^2(x)+\cos^2(x)$ is a constant, and since $\sin(0)=0$ and $\cos(0)=1$, we have
$$
\sin^2(x)+\cos^2(x)=1\tag{5}
$$
Thus, $(\cos(x),\sin(x))$ is on the unit circle and moves with speed
$$
\begin{align}
\left|\frac{\mathrm{d}}{\mathrm{d}x}(\cos(x),\sin(x))\right|
&=\left|\vphantom{\frac12}(-\sin(x),\cos(x))\right|\\[6pt]
&=1\tag{6}
\end{align}
$$
To go around the unit circle once at speed $1$ takes $2\pi$. Thus,
$$
\begin{align}
(\cos(2\pi),\sin(2\pi))
&=(\cos(0),\sin(0))\\[6pt]
&=(1,0)\tag{7}
\end{align}
$$

Combining $(1)$ and $(7)$ yields
$$
\begin{align}
\sin(x+2\pi)
&=\sin(x)\cos(2\pi)+\cos(x)\sin(2\pi)\\[6pt]
&=\sin(x)\tag{8}
\end{align}
$$
</s>