Proprieties of the solutions of the Cauchy problem associated to $f'' = -f$ Consider the Cauchy problem $$f'' (x) = -f(x)$$ with initial conditions $f(0) = 0$ and $f'(0) = 1$.
I know that it has one (and only one) solution $f$, namely $f = \sin x$, but my endgame here is to define $f:=\sin x$ and $f' := \cos x$. 
What I need to prove is the following: 

Let $f$ be the unique solution of the Cauchy problem. Then 
  
  
*
  
*$f^2(x) + f'^2(x) = 1$;
  
*$f(x+y) = f(x)f'(y) + f(y)f'(x)$;
  
*$f'(x+y) = f'(x)f'(y) - f(x)f(y)$;
  
*there exists $\epsilon > 0$ such that for all $x \in (0,\epsilon)$ we have $0 < f(x) < x < \frac{f(x)}{f'(x)}$.
  

How do I do this?
 A: By multiplying both sides of (1) by $f'(x)$, one obtains
$$ f(x)f'(x)+f'(x)f''(x)=0 $$
or
$$ \frac{d}{dx}[f^2(x)+f'^2(x)]=0 $$
and hence 
$$ f^2(x)+f'^2(x)=C $$
By using the initial conditions, one can have $C=1$ and so Part 1 is done.
For Part 2, let
$$ F(x)=f(x+y)-f(x)f'(y) - f(y)f'(x) $$
and then
\begin{eqnarray}
F'(x)&=&f'(x+y)-f'(x)f'(y) - f(y)f''(x)=f'(x+y)-f'(x)f'(y) + f(x)f(y),\\
F''(x)&=&f''(x+y)-f''(x)f'(y) + f'(x)f(y)=-f(x+y)+f(x)f'(y) +f'(x)f(y).
\end{eqnarray}
So $F(x)$ is a solution of the DE
$$ F''(x)=-F(x) $$
Noting that $F(0)=0,F'(0)=0$, by the Existence and Uniqueness Theorem of DEs, one has $F(x)\equiv0$, namely
$$ f(x+y)=f(x)f'(y)+f(y)f'(x). $$
Repeating the above, it is easy to get Part 3. Note $f'''(x)=-f'(x)$ and $f''(0)=f(0)=0$. Since $f'(0)=1$, there is $\epsilon>0$ such that in $(0,\epsilon)$, $f'(x)>0$. Now using the Taylor expansion of $f(x)$ in $(0,\epsilon)$ at $x=0$ up to order 3, one has
$$ f(x)=f(0)+f'(0)x+\frac12f''(0)x^2+\frac16f'''(c)x^3=x-\frac16f'(c)x^3<x, c\in(0,\epsilon).$$
Let $g(x)=f(x)-xf'(x)$. Then $g(0)=0,g'(0)=0,g''(x)=xf'(x)$ and
$$ g(x)=g(0)+g'(0)x+\frac12g''(c)x^2=\frac12g''(c)x^2>0,c\in(0,\epsilon) $$
which implies $f(x)>xf'(x)$ or $x<\frac{f(x)}{f'(x)}$. So Part 4 is done. 
