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