Answer
$\Sigma_{n=0}^\infty (-1)^n[\dfrac{x^{(2n)}}{(2n)!}-\dfrac{x^{(2n+1)}}{(2n+1)!}]$
Work Step by Step
Solve $\cos x-\sin x$
Then, we have $\cos x-\sin x=(\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{(2n)}}{(2n)!}-\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{(2n+1)}}{(2n+1)!})$
or, $(1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+...)-(x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...)=1-x-\dfrac{1}{2!}x^2+\dfrac{1}{3!}x^3+..=\Sigma_{n=0}^\infty (-1)^n[\dfrac{x^{(2n)}}{(2n)!}-\dfrac{x^{(2n+1)}}{(2n+1)!}]$
