We will derive the addition formulas of the sine and the cosine functions supposing known only their derivatives and the chain rule.

Define the function $F:\,\mathbb{R} \to \mathbb{R}$ through $$F(x) \;:=\; [\sin{x}\cos\alpha+\cos{x}\sin\alpha-\sin(x\!+\!\alpha)]^2+[\cos{x}\cos\alpha-\sin{x}\sin\alpha-\cos(x\!+\!\alpha)]^2$$ where $\alpha$ is, for the moment, a constant. The derivative of $F$ is easily calculated:
$F'(x) \;=\;\\ \mbox{\;\;}2[\sin{x}\cos\alpha+\cos{x}\sin\alpha-\sin(x\!+\!\alpha)][\cos{x}\cos\alpha-\sin{x}\sin\alpha-\cos(x\!+\!\alpha)]\\ +2[\cos{x}\cos\alpha-\sin{x}\sin\alpha-\cos(x\!+\!\alpha)][-\sin{x}\cos\alpha-\cos{x}\sin\alpha+\sin(x\!+\!\alpha)]$

But this expression is identically 0. By the fundamental theorem of integral calculus, $F$ must be a constant function. Since $F(0) = 0$ , we have $$F(x) \;\equiv\; 0$$ for any $x$ and naturally also for any $\alpha$ . Because $F(x)$ is a sum of two squares, the both addends of it have to vanish identically, which yields the equalities $$\sin{x}\cos\alpha+\cos{x}\sin\alpha-\sin(x\!+\!\alpha) \;=\; 0, \qquad \cos{x}\cos\alpha-\sin{x}\sin\alpha-\cos(x\!+\!\alpha) \;=\; 0.$$ These contain the addition formulas $$\sin(x\!+\!\alpha) \;=\; \sin{x}\cos\alpha+\cos{x}\sin\alpha,$$ $$\cos(x\!+\!\alpha) \;=\; \cos{x}\cos\alpha-\sin{x}\sin\alpha.$$