## University Calculus: Early Transcendentals (3rd Edition)

$1-x-\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+..=\Sigma_{n=0}^\infty (-1)^n[\dfrac{x^{2n}}{(2n)!}-\dfrac{x^{2n+1}}{(2n+1)!}]$
We know that the Maclaurin Series for $\sin x$ and $\cos x$ is defined as: $\sin x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{(2n+1)!}=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...$ and $\cos x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n}}{(2n)!}=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+...$ Then $\cos x-\sin x=(\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n}}{(2n)!})-(\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{(2n+1)!})=(1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+...)-(x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...)$ or, $1-x-\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+..=\Sigma_{n=0}^\infty (-1)^n[\dfrac{x^{2n}}{(2n)!}-\dfrac{x^{2n+1}}{(2n+1)!}]$