Given two functions $f$ and $g$ whose derivatives $f'$ and $g'$ satisfy : $f'(x)=g(x), g'(x)=-f(x),f(0)=0, g(0)=1$ for all $x$ in an interval $J$. $~~~\cdots(A)$

(a) Prove that $f^2(x)+g^2(x)=1 ~\forall~x \in J $

(b) Let $F$ and $G$ be another pair of functions in $J$ which satisfy the given conditions $(A)$. Prove that $f(x)=F(x)$ and $g(x)=G(x)$.

**Attempt:** $(a)$

$f^2(x)+g^2(x)+c= 2[ ~\int f(x) ~d ~(f(x)) + ~\int g(x) ~d ~(g(x))~]$

$=2 [ ~\int f(x) ~ g(x)~ dx - ~\int g(x) ~f(x)~ dx~]$

$ = 0$

Applying the initial conditions, we get :$f^2(x)+g^2(x)=1$

$(b)$

Consider : $h(x)=[f(x)-F(x)]^2+[g(x)-G(x)]^2$

If we prove that $h(x)=0~\forall~x \in J$, then $f(x)=F(x), g(x)=G(x)$

Expanding : $h(x)=f^2(x)+F^2(x)-2f(x)F(x)+g^2(x)+G^2(x)-2g(x)G(x)$

$=2[1-f(x)F(x)-g(x)G(x)]$

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> How do I prove that $f(x)F(x)+g(x)G(x) =1$?
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Thank you for your help.