Consider the problem $f''(x)=-f(x)$ $\forall x\in\mathbb{R}$ such that $f(0)=0$ and $f'(0)=1$.
Show that the above problem has at most one solution (assume there is a solution, and call it $s$, put $c=s'$, and use this to prove that $p=min\{x>0:c(x)=0 \}$ exists.
Showing that this has one solution is easy; further, it's even easy to show that several properties of sine and cosine are true (without any trigonometry), such as the fact that sine is odd, cosine is even, they're both bounded by one, the sum of their squares equals one, the sum and difference of angles formula, etc; all of these are easily shown by taking derivatives of some auxiliary functions. So I could potentially use any of those properties if need be. However, even with all of that information, I still struggle to show the last part.
Essentially, we need to somehow prove that this $p$, which is basically just $\pi/2$, exists. I suspect that the completeness axiom, continuity, intermediate value theorem, etc. will come into play here, but I don't really know how to formulate a proof like this. The Taylor series expansion is not being used as the definition for sine and cosine in this case, so we just have to go from what's above. How can the existence of $p$ be shown?
If $c(x_0)\le0$ for some $x_0>0$, then $c(x_1)=0$ for some $x_1>0$ (IVT). Also there is a least $x_2>0$ such that $c(x_2)=0$ (take infimum of all possible $x_1$ and apply continuity). If the conclusion fails then $c(x)>0$ for all $x>0$.
Suppose $c(x)>0$ for all $x>0$. Integrating $c$ gives $s(x)>0$ for all $x>0$. Also $c(2x)=c(x)^2-s(x)^2$ and $s(2x)=2s(x)c(x)$. You can prove this by observing $2s(x/2)c(x/2)$ satisfies the same differential equation and initial conditions as $s$, etc. Define $t(x)=s(x)/c(x)$. This is well-defined for all $x\ge0$ by our hypothesis, and $t(2x)=2t(x)/(1-t(x)^2)$. As $t(2x)>0$, then $t(x)<1$ for all $x>0$. But $t'(x)=1+t(x)^2$. Integrating gives $t(x)\ge x$ for all $x>0$. Then both $t(1)<1$ and $t(1)\ge1$.