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.
 A: 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 $f(x)=f(0)=1$.
A: Both power series are even.
The coefficient of $x^{2n}$ in the first power series is
$$
\sum_{k=0}^{n}\frac{1}{(2k-1)!(2n-2k+1)!}
$$
The coefficient of $x^{2n}$ in the second power series is
$$
\sum_{k=0}^n \frac{1}{(2k)!(2n-2k)!}
$$
So the difference between the two coefficients is
$$
\sum_{k=0}^{2n}(-1)^k \frac{1}{k!(2n-k)!}=\frac{1}{(2n)!}(1-1)^{2n} \, .
$$
A: When I ask Mathematica for the series expansion for Sin[x]^2, I get:
x^2 - x^4/3 + (2 x^6)/45 - x^8/315 + (2 x^10)/14175-...
And Cos[x]^2 is:
1 - x^2 + x^4/3 - (2 x^6)/45 + x^8/315 - (2 x^10)/14175+...
So add those two expressions together, and you're left with 1.
I'm too lazy to do it, but clearly, if you manually expanded your squared terms, you ought to be left with closed form expressions for the expansions I just listed, with the immediate consequence that added together they equal 1.
