 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 $m=n-1$ giveThe odd series similarly gives $g^{\prime }(x)={\sum }_{n=0}^{\infty }(-1{)}^n{x}^{2n}/(2n)!=f(x)$. Thus $\boxed{f^{\prime }=-g,g^{\prime }=f}$ everywhere. By the product rule,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 $\boxed{f(x{)}^2+g(x{)}^2=1\text{ for every }x\in \mathbb{R}}$. This derives the identity from the power series without assuming a trigonometric identity in advance.