Chapter 11 - Section 11.10 - Taylor and Maclaurin Series - 11.10 Exercises - Page 771: 43

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

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