Thomas' Calculus 13th Edition

$\Sigma_{n=0}^\infty (-1)^n \dfrac{(\pi)^{(2n+1)} (x)^{(2n+1)}}{(2n+1)!}$
Consider the Taylor Series for $\sin x$ as follows: $\sin x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{(2n+1)!}=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...$ Plug $(\pi x)$ instead of $x$. Then, we have $\sin (\pi x)=\Sigma_{n=0}^\infty (-1)^n \dfrac{(\pi x)^{2n+1}}{(2n+1)!}=(\pi x)-\dfrac{(\pi x)^3}{3!}+\dfrac{(\pi x)^5}{5!}-\dfrac{(\pi x)^7}{7!}+...$ Hence, $\sin (\pi x)=\Sigma_{n=0}^\infty (-1)^n \dfrac{(\pi)^{(2n+1)} (x)^{(2n+1)}}{(2n+1)!}$