Example of Parseval's Theorem

Question

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?

Answer 1

$\sin{\pi n} = 0$ when $n \in \mathbb{Z}$. Thus, because $\cos{\pi n} = (-1)^n$, we have

$$\sin{\pi (\zeta - n)} = \sin{\pi \zeta} \cos{\pi n} - \sin{\pi n} \cos{\pi \zeta} = (-1)^n \sin{\pi \zeta} $$

Answer 2

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}