There are a great many ways to prove that the derivative of sine is cosine, some of them based on things like the Taylor series definition. I’d like to prove it using only the right-triangle definition of trigonometric functions. But all the proofs I’ve seen involve the notion of the arclength of a circular arc subtended by a given angle and/or the notion of the area of a sector subtended by a given angle.
My question is, is it possible to prove that $\lim_{x\to0}\frac{\sin(x)}{x}=1$ and hence $\frac{d}{dx}\sin(x) =\cos(x)$ without invoking the notions of the arclength of a circular arc or the area of a sector? After all, it is perfectly possible to define the measure of an angle without referring to arclength; it’s done in most modern axiomatizations of Euclidean geometry, like this one:
Postulate $11$.
Angle Measurement Postulate. To every angle there corresponds a real number between $0^\circ$ and $180^\circ$.
Postulate $12$.
Angle Construction Postulate. Let $AB$ be a ray on the edge of the half-plane $H$. For every $r$ between $0$ and $180$ there is exactly one ray $AP$, with $P$ in $H$ such that $m \angle PAB = r$.
Postulate $13$.
Angle Addition Postulate. If $D$ is a point in the interior of $\angle BAC$, then $m \angle BAC = m \angle BAD + m \angle DAC$.
Postulate $14$.
Supplement Postulate. If two angles form a linear pair, then they are supplementary.
So using axioms like these, is it possible to prove that $\sin(x)\leq x \leq \tan(x)$, which is what’s needed to prove that $\lim_{x\rightarrow 0}\frac{\sin(x)}{x}=1$ and hence $\frac{d}{dx}\sin(x) =\cos(x)$?