## Thomas' Calculus 13th Edition

$\Sigma_{n=1}^\infty (-1)^{(n+1)} \dfrac{2^{(2n-1)}x^{(2n)}}{(2n)!}$
Consider the Maclaurin Series for $\cos x$ is defined as: $\cos x=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+...$ Use formula such as: $1-\cos 2x=2 \sin^2 x$ Then, we have $\sin^2 x=\dfrac{1}{2}-\dfrac{\cos 2x}{2}=\dfrac{1}{2}-\dfrac{1}{2}[1-\dfrac{(2x)^2}{2!}+\dfrac{(2x)^4}{4!}-\dfrac{(2x)^6}{6!}+...]$ $\implies \Sigma_{n=1}^\infty (-1)^{(n+1)} \dfrac{2^{(2n-1)}x^{(2n)}}{(2n)!}$