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Given the expressions:

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

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

the exercise is to show that $[f(x)]^2+[g(x)]^2$ is constant for all real $x$ and determine its value, knowing that $f$ and $g$ are real differentiable non-constant functions, and that $f'(0)=0$.

I realized it looks like the $sin$ and $cos$ functions, so the answer must be $1$. To prove something that way, I tried showing that $f$ and $g$ were always on the interval $[-1,1]$.

I have also tried to derivate each expression and plug in $x=y=0$ or only $y=0$, but was unable to develop the solution.

If we let $phi(x)=f(x)+ig(x)$, then $phi$ satisfies a functional equation
$$
phi(x+y)=phi(x)phi(y).
$$ This gives $phi(x)=phi(x)phi(0)$, so either $phi equiv 0$ or $phi(0)=1$. Since $phi equiv 0$ is a trivial solution, we assume $phi(0)=1$. Differentiating with $y$ and plugging $y=0$, we obtain $phi'(x)=phi'(0)phi(x)$. Now, define $u(x)=e^-phi'(0)xphi(x)$ and observe that
$$
u'(x)=e^-phi'(0)xleft(phi'(x)-phi'(0)phi(x)right)=0.
$$ This gives $u(x) = u(0)=phi(0)=1$ and hence $phi(x) = e^phi'(0)x$. Since $$phi'(0)=f'(0)+ig'(0)=ig'(0)$$ it follows $phi(x) = e^ig'(0)x=e^itheta x$ for some $thetainBbb R$, hence by Euler's identity $$
left(f(x)right)^2+left(g(x)right)^2 =cos^2(theta x)+sin^2(theta x) =1.
$$ (Also note that trivial solution $phi =0$ gives $left(f(x)right)^2+left(g(x)right)^2 =0$.)

Let's derive your two equations, with respect to $x$ : you get
$$f'(x+y)=f'(x)f(y)-g'(x)g(y)$$
and
$$g'(x+y)=f'(x)g(y)+f(y)g'(x)$$

Evaluating them in $x=0$ gives you, because $f'(0)=0$,
$$f'(y)=-g'(0)g(y)$$
and
$$g'(y)=f(y)g'(0)$$

So for all $y$,
$$f(y)f'(y) + g(y)g'(y) = - g'(0)g(y)f(y)+f(y)g'(0)g(y) = 0$$

You get that
$$2(f(y)f'(y) + g(y)g'(y))=0, quad texti.e. (f^2+g^2)'(y)=0$$

So $f^2 + g^2$ is constant.

To find its value, plug $x=y=0$ in the two original equalities : you get
$$f(0)=f^2(0)-g^2(0)$$
and
$$g(0)=2f(0)g(0)$$

If $g(0) neq 0$, you would get from the second equation $2f(0)=1$, so $f(0)=frac12$. The first equation qives you $g^2(0)= frac14-frac12 = -frac12$, which is impossible.

So $g(0)=0$, so the first equation gives you $f(0)=f^2(0)$, so $f(0)=0$ or $1$. But $f(0)=0$ is impossible because plugging $y=0$ in $f'(x+y)=f'(x)f(y)-g'(x)g(y)$ would say that $f'(x)=0$ and $f$ is constant, which is impossible. So $f(0)=1$, so you deduce that $f^2(0)+g^2(0)=1$.

Finally the value of the constant $f^2 + g^2$ is $1$.