Prove that sin and cos are the unique functions satisfying given properties

Let $f, g$ be given functions with derivatives $f', g'$ satisfying

$f'(x) = g(x), \quad \text{and} \quad g'(x) = -f(x), \quad \text{and} \quad f(0) = 0, \quad \text{and} \quad g(0) = 1,$

for all $x$ in some interval $J$ which contains 0. (Note: $f(x) = \sin x$ and $g(x) = \cos x$ satisfy all of these conditions.)

Prove that these functions must satisfy the Pythagorean identity
$f^2(x) + g^2(x) = 1 \qquad \text{for all } x \in J.$
If $F(x)$ and $G(x)$ are any other pair of functions satisfying all of these conditions prove that
$F(x) = f(x) \qquad \text{and} \qquad G(x) = g(x) \qquad \text{for all } x \in J.$
What else can we say regarding the functions $f$ and $g$ ?

Proof. First, we show that $f^2(x) + g^2(x) = c$ for some constant, and then we will prove that constant must be 1. To show $f^2 + g^2$ is constant we take the derivative
$\begin{align*} \left( f^2(x) + g^2(x) \right)' &= \left( f^2(x) \right)' + \left( g^2(x) \right)' \\ &= 2 f(x) f'(x) + 2 g(x)g'(x) \\ &= 2 f(x) g(x) - 2 f(x) g(x) &(f'(x) = g(x), \ g'(x) = -f(x)) \\ &= 0. \end{align*}$

This holds for all $x \in J$ ; hence, by the zero derivative theorem (Theorem 4.7 (c) in Apostol) we must have $f^2(x) + g^2(x) = c$ for some constant $c$ for all $x \in J$ . Now, to show this constant $c = 1$ , we know by hypothesis that $0 \in J$ and we have

$f^2(0) + g^2(0) = 0 + 1 = 1.$

Hence, $f^2 (x) + g^2(x) = 1$ everywhere on $J. \qquad \blacksquare$
Proof. Let $F(x)$ and $G(x)$ be another pair of functions satisfying the given relations:
$F'(x) = G(x), \qquad G'(x) = -F(x), \qquad F(0) = 0, \qquad G(0) = 1.$

Then define

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

This implies

$h'(x) = 2(F(x) - f(x))(F'(x) - f'(x)) + 2(G(x) - g(x))(G'(x) - g'(x)).$

Using the relations $G'(x) = -F(x), \ g'(x) = -f(x), \ G(x) = F'(x)$ , and $g(x) = f'(x)$ we have

$\begin{align*} h'(x) &= 2(F(x) - f(x))(F'(x) - f'(x)) + 2(F'(x) - f'(x))(f(x) - F(x))\\ &= 2(F(x) - f(x))(F'(x) - f'(x)) - 2(F(x) - f(x))(F'(x) - f'(x)) \\ &= 0 \end{align*}$

for all $x \in J$ . Hence, $h(x) = c$ for some constant $c$ . Then, we evaluate $h(0)$ and use the given relations $f(0) = F(0) = 0$ and $g(0) = G(0) = 1$ ,

$\begin{align*} h(0) = c &= (F(0) - f(0))^2 + (G(0) - g(0))^2 \\ &= 0 + (1-1)^2 \\ &= 0. \end{align*}$

Hence, $h(x) = 0$ for all $x \in J$ . Since it is a sum of squares (which must be nonnegative) we have that $h(x) = 0$ if and only if

$F(x) - f(x) = 0 \qquad \text{and} \qquad G(x) - g(x) = 0$

for all $x \in J$ . Hence we must have

$F(x) = f(x) \qquad \text{and} \qquad G(x) = g(x). \qquad \blacksquare$
I’m not entirely sure what Apostol wants us to say here with respect to this solution. Since we have established that $\sin x$ and $\cos x$ satisfy these properties and that any functions satisfying these properties are unique, we can conclude that $\sin x$ and $\cos x$ are the unique functions which satisfy the given properties.