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 $g(x)=-g(-x)$
Since the proof comes directly from Kelenner's comment, I post this answer in community wiki, so as to not earn undeserved points.

We assume $f(0)=1$ and

$$\forall x,y \in \Bbb R, f(x-y)=f(x)f(y)-g(x)g(y)$$

Notice that we use nowhere the assumption that $f$ or $g$ is continuous.

Then, for all $x\in\Bbb R$,

$$1=f(0)=f(x-x)=f(x)^2-g(x)^2$$

And

$$f(2x)=f(x)f(-x)-g(x)g(-x)$$

Also, $\forall x,y\in \Bbb R$,

$$f(y-x)=f(x)f(y)-g(x)g(y)=f(x-y)$$

Hence $\forall x\in\Bbb R$, $f(x)=f(-x)$, that is, $f$ is even.

Therefore, $f(-x-y)=f(x+y)$, or, $\forall x,y\in \Bbb R$,

$$f(-x)f(y)-g(-x)g(y)=f(x)f(-y)-g(x)g(-y)$$

Hence, $\forall x,y\in \Bbb R$, $g(-x)g(y)=g(x)g(-y)$.

Now, let's suppose $g(y_0)\neq0$ for some $y_0 \in \Bbb R$.

We have then, $\forall x\in\Bbb R$, such that $g(x)\neq0$,
$$\frac{g(-x)}{g(x)}=\frac{g(-y_0)}{g(y_0)}=a$$

Where $a$ is a constant real number.

And if $g(x)=0$, then from $g(-x)g(y_0)=g(x)g(-y_0)=0$ and $g(y_0)\neq0$, we have that $g(-x)=0$ too.

That is, $\forall x\in\Bbb R$, $g(-x)=ag(x)$.

But then $\forall x\in\Bbb R$, $g(x)=ag(-x)=a^2g(x)$, and since $g(y_0)\neq0$, we can simplify and $a^2=1$, or $a=\pm1$. Thus the function $g$ is either even or odd.

If $g$ is even, then $\forall x\in\Bbb R$,

$$f(2x)=f(x)^2-g(x)^2=1$$

So $f$ is constant ($=1$).

But then $\forall x\in\Bbb R$, $g(x)^2=f(x)^2-f(2x)=0$.

Hence, if $g$ is even, it's the null function.

But that means $g$ is always odd, since either it is odd, either it is null (thus also odd).
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 $\mathbb{R}=\Gamma^-$.