How to compute $\sin{(\pi x)}$ about $\frac12$ in taylor series? The correct answer is supposed to be $\sum\frac{(-1)^n}{(2n)!}\pi^{2n}(x-\frac12)^n$ which I don't understand. Since the function is about $x=\frac12$, so $(x-\frac12)^n$ is good. But $\sum\frac{(-1)^n}{(2n)!}\pi^{2n}(x-\frac12)^n$ is supposed to be the Taylor expansion for $\cos(\pi(x-\frac12))$. How do you get such answer?

$\sin \pi x=\sin [\pi (x-1/2)+\pi /2]=\cos [\pi (x-1/2)],$ because $\sin (z+\pi /2)=\cos z.$
