Example of Parseval's Theorem In the textbook "Mathematics for Physics" of Stone and Goldbart the following example for an illustration of Parseval's Theorem is given:


Until 2.42 I understand everything but I don't understand the statement:
" Finally, as $\sin^2(\pi(\zeta-n))=\sin^2(\pi \zeta)$ "
Can you explain me why this equality holds?

Let's work it out from first principles. Recall that
$$\sin(a+b)=\cos(b)\sin(a)+\cos(a)\sin(b)$$
Now, we consider $\sin(\pi(\zeta-n))=\sin(\pi\zeta-\pi n)$, i.e. $a=\pi\zeta$ and $b=-\pi n$:
\begin{align}\sin(\pi\zeta-\pi n)&=\cos(-\pi n)\sin(\pi\zeta)+\cos(\pi\zeta)\sin(-\pi n)\\
&=\cos(\pi n)\sin(\pi\zeta)-\cos(\pi\zeta)\sin(\pi n)\end{align}
where we have used that:
\begin{align}\cos(-\theta)&=\cos(\theta)\\\sin(-\theta)&=-\sin(\theta)\end{align}
And by periodicity (since $n$ is an integer):
\begin{align}\cos(\pi n)&=\pm 1\\\sin(\pi n)&=0\end{align}
Therefore:
\begin{align}\sin(\pi\zeta-\pi n)&=\pm\sin(\pi\zeta)\\
\sin^2(\pi\zeta-\pi n)&=\left[\pm\sin(\pi\zeta)\right]^2\\
&=\sin^2(\pi\zeta)\end{align}
