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Let $R_f,R_g$ be the two radii of convergence. First take $|x|<t<R_f$, with $t>0$. Absolute convergence at $t$ bounds the terms: $|a_n{t}^n|\le M$ for all $n$. ThusAt $x=0$ the right-hand series is finite. Otherwise it converges by the ratio test, because its limiting ratio is $|x|/t<1$. So the differentiated power series converges absolutely at every interior point of $f$, proving $R_g\ge R_f$. This proof also covers $R_f=\infty$ by choosing a finite $t$ above any given $|x|$; if $R_f=0$, that inequality is automatic.

Conversely, if $0<|x|<R_g$, absolute convergence givesThe constant term does not affect convergence, and both series converge at zero. Hence $R_f\ge R_g$, including the zero and infinite cases. Therefore $\boxed{R_f=R_g}$. This proof uses coefficient estimates, rather than assuming the termwise differentiation of a power series result whose radius assertion is being established.

The termwise differentiation of a power series theorem states that inside its radius of convergence, the sum is differentiable and its derivative is the sum of the differentiated series. Accordingly $\boxed{f^{\prime }(x)=g(x)\text{ for }|x|<R}$. There is no blanket differentiability assertion at the convergence endpoints.

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.