Answer
$\Sigma_{n=0}^{\infty}(-1)^{n}\dfrac{\pi ^{(2n+1)}x^{(2n+1)}}{(2n+1)!}$
Work Step by Step
$sin\pi x=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{(\pi x)^{(2n+1)}}{(2n+1)!}$
$=\Sigma_{n=0}^{\infty}(-1)^{n}\dfrac{\pi ^{(2n+1)}x^{(2n+1)}}{(2n+1)!}$