Answer
$\dfrac{1}{5!}x^5-\dfrac{1}{7!}x^7+\dfrac{1}{9!}x^9-\dfrac{1}{11!}x^{11}+..$
Work Step by Step
Consider the Maclaurin Series for $\cos x$ is defined as:
$ \sin x=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...$
Then, $\sin x-x+\dfrac{x^3}{3!}=(x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...)-x+\dfrac{x^3}{3!}$
$\implies \dfrac{1}{5!}x^5-\dfrac{1}{7!}x^7+\dfrac{1}{9!}x^9-\dfrac{1}{11!}x^{11}+..$
