Even function about a point over a restricted range

Question

Why is $f(x)=(x-1)^2$sin$(n\pi x)$ even about $x=1$ for $0\leq x \leq2$?

I understand that $(x-1)^2$ is even about $x=1$ and I can plot the graph for various values of $n$ on wolfram alpha, but how do I go about proving the above is correct?

Answer 1

$f(1+a)=a^2\sin(n\pi(1+a))=a^2\sin(n\pi+n\pi a)=a^2(\sin n\pi\cos n\pi a+\cos n\pi\sin n\pi a)$

$=a^2\cos n\pi\sin n\pi a$, but

$f(1-a)=a^2\sin(n\pi(1-a))=a^2\sin(n\pi-n\pi a)=a^2(\sin n\pi\cos n\pi a-\cos n\pi\sin n\pi a)$

$=-a^2\cos n\pi\sin n\pi a$; so the function does not appear to be even about 1.

For example, if $n=1$, $f(3/2)=-1/4$ while $f(1/2)=1/4$.