The problem:
Defining $\sin x$ as the leg $b$ of a right triangle with $\angle B=x$ (in radians) and hypotenuse $1$, prove that $$\lim_{x\to 0}\frac{\sin x}x=1$$
(The motivation is to find the derivative of $\sin x$ in a elementary, "pre-Taylor" and "pre-series" context).
I have seen many times a proof that it is based in the fact that the length of the arc $x$ satisfies $$\sin x<x<\tan x$$ or sometimes $$\sin x<x<\sin x+1-\cos x$$ The lower bound is clear because the $\sin x$ is the length of a straight segment and $x$ is the length of a curved segment with the same endpoints. But I find that the upper bound is based on this intuitive fact:
If $F$ and $G$ are two convex subsets of $\Bbb R^2$ and $F\subset G$, then $|\partial F|<|\partial G|$.
($|\partial F|$ is the length of the boundary of $F$).
But I haven't ever seen a proof of that. I tried it myself but I get stuck trying to bound $$\int_s^t\sqrt{1+y'(u)^2}du$$ provided for example that $y''$ is negative and $y(s)=y(t)$. I realize that $y'$ can't be bounded (note for example $y=\sqrt{1-x^2}$, $-1\le x\le 1$).
Question
Is it possible to justify the derivative of $\sin x$ or the inequality about the lengths of bounds? Is there any calculus text with this approach (or similar) to define trigonometical functions?
The first inequality is easier to justify by comparing areas(How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?), but I suppose the more interesting question is the inequality
$$\text{length}(y_1) \leq\text{length}{(y_2)}$$ under the assumptions that the graphs $y_i$ are concave, $y_1\le y_2$, and $y_1=y_2$ at the endpoints. You can prove it first for polygonal graphs, and then approximate the continuous graph by a sufficiently detailed polygonal graph.
For polygonal paths, you can proceed as follows. The result is obvious(triangle inequality) if $y_1$ is flat. Otherwise, extend the 2nd segment of $y_1$ backwards until you hit $y_2$ at say $X$. The resulting path obtained from going along $y_1$ until $X$, then switching to $y_1$ is strictly longer than $y_1$, by triangle inequality a large number of times. Now we can ignore the curves up to the point $X$, so we have reduced the number $n$ of segments of $y_1$. The result follows from induction in $n$.
A visual aid -