Maclaurin series: see reply for question (couldnt fit)

Question

anonymous · Answer

I am working on the Maclaurin series of $$f(x)=sin^2(x)$$ First I've written it as: $$f(x)=sin^2(x)=\frac{1}{2}(1-cos(2x))$$ So I first looked at the maclaurin series for $$cos(2x)=1-\frac{(2x)^2}{2!}+\frac{(2x)^4}{4!}-\frac{(2x)^6}{6!}+...$$Now I look at $$\frac{1}{2}(1-cos(2x))=\frac{1}{2}(1-(1-\frac{(2x)^2}{2!}+\frac{(2x)^4}{4!}-\frac{(2x)^6}{6!}+...))$$
$$=\frac{1}{2}(1-1+\frac{(2x)^2}{2!}-\frac{(2x)^4}{4!}+\frac{(2x)^6}{6!}-...)$$ $$=\frac{1}{2}-\frac{1}{2}+\frac{1}{2}\frac{(2x)^2}{2!}-\frac{1}{2}\frac{(2x)^4}{4!}+\frac{1}{2}\frac{(2x)^6}{6!}-...$$
Would this be correct? And how cold I write this as a sum?

zarkon · Answer

$$\sum_{k=1}^{\infty}(-1)^{k+1}\frac{(2x)^{2k}}{2(2k)!}$$

anonymous · Answer

So my intermediate calculations are right?

zarkon · Answer

yes

anonymous · Answer

Okay, thanks :)