Interesting sine/cosine problem about zeros

Question

Suppose we have two differentiable functions $f, g:\mathbb{R}\to\mathbb{R}$ such that $f(0)=0$, $g(0)=1$, $f'(x)=g(x)$ and $g'(x)=-f(x)$. Then it is possible to show not only that $f(x)=\sin(x)$ and $g(x)=\cos(x)$ satisfy these conditions, but that they are the only pair of functions that satisfy this condition. For instance, suppose $(f_1, g_1)$ and $(f_2, g_2)$ are two pairs of functions that satisfy this equation, then $$\frac{d}{dx}[(f_1-f_2)^2+(g_1-g_2)^2]=2(f_1-f_2)(f_1'-f_2')+2(g_1-g_2)(g_1'-g_2')$$ which is zero. This implies this function is constant, and is equal to zero by substituting $x=0$. This implies $f_1=f_2$ and $g_1=g_2$. In light of this, we can say a definition of $\sin(x)$ and $\cos(x)$ are two functions that satisfy these conditions. My question is whether we can show that $\sin(\pi)=0$ and $\cos(\pi/2)=0$ just from these conditions. Stated more formally,

Suppose we have two differentiable functions $f, g:\mathbb{R}\to\mathbb{R}$ such that $f(0)=0$, $g(0)=1$, $f'(x)=g(x)$ and $g'(x)=-f(x)$. Can I show that $f(\pi)=0$ and $g(\pi/2)=0$?

Answer 1

Yes you can. From the equations \begin{align*} f'(x) &= g(x) \\ g'(x) &= -f(x) \end{align*} you may perform the substitution $\alpha = g + if$ and $\beta = g-if$ to obtain the following: \begin{align*} \alpha'(x) &= g'(x) + if'(x) = -f(x) + i g(x) =i\alpha(x) \\ \beta'(x) &= g'(x) -i f'(x) = -f(x) - i g(x) = -i\beta(x) \end{align*} The solutions to ODEs taking the form $\phi'(x) = k\phi(x)$ are known to take the form $\phi(x) = Ce^{kt}$, hence we may conclude that \begin{align*} \alpha(x) &= C_1 e^{ix} \\ \beta(x) &= C_2 e^{-ix} \end{align*} for some constants $C_1,C_2\in\mathbb C$. Using the initial conditions $(f,g)(0) = (0,1)$, you can find that $(\alpha,\beta)(0) = (1,1)$, and therefore $C_1 = C_2 = 1$. This means that \begin{align*} g + if &= e^{ix} \\ g - if &= e^{-ix} \end{align*} From here, you can solve for $f,g$ in terms of $x$.