Answer
$x-\dfrac{4}{3!}x^3+\dfrac{16}{5!}x^5-\dfrac{64}{7!}x^7+...=\Sigma_{n=0}^\infty (-1)^n\dfrac{2^{(2n)}x^{(2n+1)}}{(2n+1)!}$
Work Step by Step
Consider the Maclaurin Series for $\sin x$ is defined as:
$ \sin x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{(2n+1)!}=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...$
Use formula such as: $\sin x \cos x=\dfrac{\sin 2x}{2}$
Then, we have $\sin x \cos x=\dfrac{\sin 2x}{2}=\dfrac{2x-\dfrac{(2x)^3}{3!}+\dfrac{(2x)^5}{5!}-\dfrac{(2x)^7}{7!}+...}{2}$
Thus, $x-\dfrac{4}{3!}x^3+\dfrac{16}{5!}x^5-\dfrac{64}{7!}x^7+...=\Sigma_{n=0}^\infty (-1)^n\dfrac{2^{(2n)}x^{(2n+1)}}{(2n+1)!}$
