Forget about the sin and cos functions, show that $(x-x^3/3!+x^5/5!-x^7/7!+...)^2+ (1-x^2/2!+x^4/4!-x^6/6!+...)^2=1$. Forget about the $\sin$ and $\cos$ functions, are there possibly some brilliant way to show that 
$$\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots\right)^2+
\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots\right)^2=1$$
?
I've thought for a long time, without making much progress. Can someone help me? Thanks.

If we denote
$$f(x)=\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots\right)^2+
\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots\right)^2$$
then by differentiation term by term we see that $f'(x)=0$ so $f(x)=f(0)=1$.
