= Solution
Since the common <radius of convergence> is infinite, the <termwise differentiation of a power series> theorem applies on the entire real line. For the even series, differentiation and the index change $m=n-1$ give
$$
f'(x)=\sum_{n=1}^\infty(-1)^n\frac{x^{2n-1}}{(2n-1)!}
=-\sum_{m=0}^\infty(-1)^m\frac{x^{2m+1}}{(2m+1)!}=-g(x).
$$
The odd series similarly gives $g'(x)=\sum_{n=0}^\infty(-1)^n x^{2n}/(2n)!=f(x)$. Thus \b[$\boxed{f'=-g,\qquad g'=f}$] everywhere. By the <product rule>,
$$
\frac d{dx}(f^2+g^2)=2ff'+2gg'=-2fg+2gf=0.
$$
The <mean value theorem> implies that a differentiable function with zero derivative is constant on the real line. At zero, the series give $f(0)=1$ and $g(0)=0$. Consequently \b[$\boxed{f(x)^2+g(x)^2=1\text{ for every }x\in\mathbb R}$]. This derives the identity from the <power series> without assuming a trigonometric identity in advance.
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