Chapter 11 - Infinite Sequences and Series - 11.10 Taylor and Maclaurin Series - 11.10 Exercises - Page 811: 36

$\Sigma_{n=0}^{\infty}(-1)^{n}\dfrac{{(\frac{\pi }{4})^{(2n+1)}}x^{2n+1}}{(2n+1)!}$

$f(x)=sin(\pi x/4)$ Since, $sinx=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{{(x)}^{(2n+1)}}{(2n+1)!}$ Change $x$ to $\pi x/4$ $sin(\frac{\pi x}{4})=\Sigma_{n=0}^{\infty}(-1)^{n}\dfrac{{(\frac{\pi x}{4})^{(2n+1)}}}{(2n+1)!}$ $=\Sigma_{n=0}^{\infty}(-1)^{n}\dfrac{{(\frac{\pi }{4})^{(2n+1)}}x^{2n+1}}{(2n+1)!}$