Could someone give me feedback regarding whether or not I'm on the right track for this proof? Thanks in advance!
Suppose that $ f,g\colon \mathbb{R}\to \mathbb{R}$ are each differentiable and that:
$\forall$ $x$ $\in \mathbb{R}$ $f'=g$.
$\forall$ $x$ $\in \mathbb{R}$ $g'=-f$.
$f(0)=0$.
$g(0)=1$.
Show that $\forall$ $x$ $\in \mathbb{R}$ $f^{2}+g^{2}=1$.
This is what I was thinking:
We know $ f,g\colon \mathbb{R}\to \mathbb{R}$ are each
differentiable. Consequently by a theorem in the book, the product $fg$ is differentiable. More specifically, $(fg)'=f'g+g'f$. Since $f'=g$ $\forall$ $x$ $\in \mathbb{R}$ and $g'=-f$ $\forall$ $x$ $\in \mathbb{R}$ we have that $(fg)'=g^{2}-f^{2}$.
And now here lies my problem. I was trying to come up with different ways that sums or products would give me $g^{2}+f^{2}$ and then I could use the final two conditions. But haven't been able to do so. Am I on the right track? Or am I not making any sense?
It makes sense, but I don't see how the prove what you want to prove starting like that.
Note that$$(f^2+g^2)'=2ff'+2gg'=2(fg-gf)=0.$$Therefore, $f^2+g^2$ is constant. But $f^2(0)+g^2(0)=1$, and so $f^2+g^2=1$.