Answer
$\Sigma_{k=0}^n \dfrac{(-1)^k \pi^{2k}}{(2k)!}(x-\dfrac{1}{2})^{2k}$
Work Step by Step
The nth Taylor polynomial for a function $f$ is
$$p_n(x)= f(x_0) + f'(x_0)(x-x_0) +\dfrac{f''(x_0)}{2!} (x-x_0)^2 +.......+\dfrac{f^n(x_0)}{n!} (x-x_0)^n$$
$f(x)=\sin \pi x$
Differentiate w.r.t x
$f'(x)=\pi cos\pi x$
$f''(x)=-\pi^2 \sin \pi x$
$f'''(x)=-\pi^3 \cos \pi x$
$f''''(x)=\pi^4 \sin \pi x$
Now $f(\dfrac{1}{2})=1$, $f'(\dfrac{1}{2})=0$, $f''(1/2)=-\pi^2$, $f'''(1/2)=0$, $f''''(1/2)=\pi^4$
Hence the nth Taylor polynomial for a function is:
$\displaystyle 1-\dfrac{\pi^2}{2}(x-\dfrac{1}{2})^2+\dfrac{\pi^4}{24}(x-\dfrac{1}{2})^4+ .......+\dfrac{(-1)^n \pi^{2n}}{(2n)!}(x-\dfrac{1}{2})^{2n}=\Sigma_{k=0}^n \dfrac{(-1)^k \pi^{2k}}{(2k)!}(x-\dfrac{1}{2})^{2k}$
