Answer
$\Sigma_{n=1}^\infty \dfrac{(-1)^{n+1} 2^{2n-1}x^{2n}}{(2n)!}$
Work Step by Step
Since, we know that the Maclaurin Series for $\cos x$ is defined as:
$ \cos x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n}}{(2n)!}=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+...$
We also know that $1-\cos 2x=2 \sin^2 x$
Now,
$ sin^2 x=\dfrac{1}{2}-\dfrac{\cos 2x}{2}=\dfrac{1}{2}-\dfrac{1}{2}[1-\dfrac{(2x)^2}{2!}+\dfrac{(2x)^4}{4!}-\dfrac{(2x)^6}{6!}+...]$
or, $=\Sigma_{n=1}^\infty \dfrac{(-1)^{n+1} 2^{2n-1}x^{2n}}{(2n)!}$
