How prove $g(x)$ is odd function :$g(x)=-g(-x)$ QUestion:

let $f(x),g(x)$ is continuous on $R$,and such
  $$f(x-y)=f(x)f(y)-g(x)g(y)$$
  and $f(0)=1$
show that: for any $x\in R$, have $g(x)=-g(-x)$

my try: let $x=y=0$,then
$$f(0)-[f(0)]^2=g(0)g(0)\Longrightarrow g(0)=0$$
and let $x=0$, note $f(0)=1,g(0)=0$,so
$$f(-y)=f(y)$$
since
$$g(x)g(y)=f(x)f(y)-f(x-y)$$
so
$$g(-x)g(y)=f(-x)f(y)-f(-x-y)=f(x)f(y)-f(x+y)$$
since
$$f(x+y)=f(x)f(-y)-g(x)g(-y)\Longrightarrow g(x)g(-y)=f(x)f(y)-f(x+y)$$
so
$$g(-x)g(y)=g(x)g(-y)$$
But I can't have $g(x)=-g(-x)$

First note that by using $f(x)=f(-x)$
$$
f(0) = f(x-x) = f(x)^2- g(x)^2  \\
f(0) = f(-x-(-x))=f(x)^2-g(-x)^2 \\
g(x)^2 = g(-x)^2 
$$
Thus let one define into two sets (note an element may be in both)
$$
\Gamma^+=\{x \in \mathbb{R} : g(x) = g(-x) \} \\
\Gamma^-=\{x \in \mathbb{R} : g(x) = -g(-x) \} \\
$$
Clearly $\mathbb{R}=\Gamma^+ \cup \Gamma^-$, such that $g$ is even for $x\in\Gamma^+$, and odd for $x\in\Gamma^-$.
Consider $x\in\Gamma^+$, one has
$$
f(x-y)=f(x)f(-y)-g(x)g(-y) \\
\therefore f(x-y)=f(x+y)\Rightarrow f(x-x)=f(x+x) \Rightarrow f(x)=1 \:\forall x\textrm{ as }f(0)=1 \\
\Rightarrow g(x)=0 \:\forall x \in\Gamma^+
$$
Thus $x\in\Gamma^+\Rightarrow g(x)=0 \Rightarrow g(x)=-g(-x)\Rightarrow x\in\Gamma^-$. Thus $\mathbb{R}=\Gamma^-$.
