use a maclaurin series to obtain the maclaurin series for the given function f(x)= (sinx)^2 hint use (sinx)^2 = 1/2(1-cos2x)
$\displaystyle f(x)=sin^2(x)=\frac{1}{2}-\frac{1}{2}cos(2x)$
Basically this amounts to finding the maclaurin series for cos(u) since then you can just put in a 2x for u (then multiply by your 1/2 scalar and subtract from 1/2). Note this trick works because the radius of convergence for this series turns out to be infinite, more care should in general be used.
let
f(u)=cos(u) f(0)=1
f'(u)=-sin(u) f'(0)=0
f"(u)=-cos(u) f"(0)=-1
f"'(u)=sin(u) f"'(0)=0
and repeat.
so:
$\displaystyle cos(u)=1 + 0u -\frac{1}{2!}u^2 + 0u^3 + \frac{1}{4!}u^4... =\sum_{n=0}^{\infty}\frac{(-1)^nu^{2n}}{2n!}$
so finally you see:
$\displaystyle sin^2(x)=\frac{1}{2}(1-cos(2x))=\frac{1}{2}\left(1-\sum_{n=0}^{\infty}\frac{(-1)^n(2x)^{2n}}{2n!}\right)$
Think you can take her from here?