Answer
$\cos \ \pi x =\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{\pi^{2n}(x)^{2n}}{(2n)!}$
Work Step by Step
We know that the Maclaurin series for $\cos x=\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{(x)^{2n}}{(2n)!}=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+..... $
Replace $x$ with $\pi x$ in the above series.
$\cos \pi x=1-\dfrac{(\pi x)^2}{2!}+\dfrac{(\pi x)^4}{4!}-\dfrac{(\pi x)^6}{6!}+..... =1-\dfrac{\pi^2 x^2}{2}+\dfrac{\pi^4 x^4}{24}-.......$
Thus, our required series is:
$\cos \ \pi x =\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{\pi^{2n}(x)^{2n}}{(2n)!}$
