Geometrically showing $\frac{\alpha}{\beta} > \frac{\sin\alpha}{\sin\beta}$, for $0 < \beta < \alpha < 90^\circ$ 
I attempting to prove the show trig identity, i.e. $\frac{\alpha}{\beta} > \frac{\sin\alpha}{\sin\beta}$. I have found that the radius is $1$ so there are three line segments equal to $1$, but I am unsure which segment would be $\frac{\sin\alpha}{\sin\beta}$?
 A: $$\sin x < x < \tan x \qquad \forall~x \in [0, \frac{\pi}2]$$
If one agrees to start with the above known fact (which is purely an intuitive geometric fact as detail in e.g. this answer and NOT about algebraic analysis of functions), then:
On one hand, we have 
$$ \frac{\alpha - \beta}2 > \sin \bigl( \frac{ \alpha -\beta}2\bigr) \quad \implies\quad \alpha - \beta > 2 \sin \bigl( \frac{ \alpha -\beta}2\bigr)  \tag*{Eq.(1)}$$
On the other hand, $ \alpha > \beta$ and cosine being a decreasing function yield
$$ \beta < \tan\beta = \frac{ \sin\beta }{ \cos\beta } < \frac{ \sin\beta }{ \cos\bigl( \frac{ \alpha + \beta}2 \bigr) } \quad \implies \quad \sin\beta > \beta \cdot \cos\bigl( \frac{ \alpha + \beta}2 \bigr)  \tag*{Eq.(2)}
$$
The product of the two inequalities (all terms are positive) gives us
\begin{align}
&& (\alpha - \beta) \cdot \sin\beta &> \beta \cdot 2 \sin\bigl( \frac{ \alpha - \beta}2 \bigr)\cos\bigl( \frac{ \alpha + \beta}2 \bigr) \\
&\implies & (\alpha - \beta) \cdot \sin\beta &> \beta \cdot (\sin\alpha - \sin\beta) \\
&\implies & \frac{\alpha - \beta}{ \beta } &> \frac{\sin\alpha - \sin\beta}{ \sin\beta}\qquad \text{, then $+1$ on both sides} \\
&\implies & \frac{ \alpha }{ \beta } &> \frac{ \sin\alpha }{ \sin\beta}
\end{align}
Note that the whole derivation can be viewed as a purely geometric proof expressed via algebra. One can make a sketch and find ALL the relevant lengths (including e.g. $\frac{ \sin\beta }{ \cos( \frac{ \alpha + \beta}2 ) }$).
A: Consider first the case where $\alpha, \beta$ are commensurable, i.e.:
$$
\alpha = n\gamma, \ \beta = m\gamma \quad (0 < m < n).
$$
By taking a smaller value of $\gamma$, if necessary, we can suppose
that $m > 1$.
Inscribe in the unit circle the broken line [polygonal chain]
with vertices:
$$
A_j = (\cos(j\gamma), \sin(j\gamma)) \quad (0 \leqslant j \leqslant n).
$$
Denoting the origin by $O$, we have:
\begin{align*}
\frac{\alpha}{\beta} = \frac{n}{m} =
& \frac{\text{area of polygon } OA_0A_1{\cdots}A_n}
{\text{area of polygon } OA_0A_1{\cdots}A_m}, \\
\frac{\sin\alpha}{\sin\beta} =
& \frac{\text{area of triangle } OA_0A_n}
{\text{area of triangle } OA_0A_m}, \\
\therefore\ 
\frac{\sin\alpha}{\sin\beta} < \frac{\alpha}{\beta}
\iff
& \frac{\text{area of triangle } OA_0A_n}
{\text{area of triangle } OA_0A_m}
< \frac{n}{m} \\
\iff
& \frac{\text{area of polygon } A_0A_1{\cdots}A_n}
{\text{area of polygon } A_0A_1{\cdots}A_m} 
> \frac{n}{m}.
\end{align*}
Define:
$$
a_j = \text{area of triangle } A_0A_jA_{j+1} \quad (0 < j < n).
$$
Then:
$$
\frac{\sin\alpha}{\sin\beta} < \frac{\alpha}{\beta}
\iff
\frac{a_1 + a_2 + \cdots + a_{n-1}}
{a_1 + a_2 + \cdots + a_{m-1}}
> \frac{n}{m}.
$$It is enough to prove that the sequence $(a_j)$ is strictly
increasing, because this implies:
\begin{align*}
\frac{a_1 + a_2 + \cdots + a_{n-1}}
{a_1 + a_2 + \cdots + a_{m-1}}
& = 1 + \frac{a_m + a_{m+1} + \cdots + a_{n-1}}
{a_1 + a_2 + \cdots + a_{m-1}} \\
& \geqslant 1 + \frac{(n - m)a_m}{(m - 1)a_{m-1}}
> 1 + \frac{n - m}{m - 1}
= \frac{n - 1}{m - 1} \\
& > \frac{n}{m}.
\end{align*}
If $B_j$ is the foot of the perpendicular from $A_0$ to $A_{j+1}A_j$
produced, then $A_0B_{j+1}$ intersects $A_jB_j$ at a point $C$, and
$A_0B_{j+1} > AC > A_0B_j$, whence $a_{j+1} > a_j$, as required.
$\square$
This figure illustrates the case $n = 5$, $m = 3$, $j = 2$:

If $\alpha, \beta$ are not commensurable, we can (as moderns!)
appeal to continuity, in order to complete the proof.  Although I
don't yet see how the ancient Greeks would have dealt with the
incommensurable case, I don't doubt that they did deal with it,
and probably with great ease.
(One might expect to find in
Euclid
a proposition to the effect that if $m < n$, then the area of a
regular $m$-gon is less than that of a regular $n$-gon inscribed
in the same circle; but it doesn't seem to be there.
This may be worth a question in History of Science and Mathematics
SE - unless someone here already knows where such a result may be
found in extant Greek mathematical texts.)
A: Two approaches

Concavity of $\boldsymbol{\sin(x)}$
If $f(x)$ is concave, then $\frac{f(x)-f(y)}{x-y}$ is a decreasing function of $x$ and a decreasing function of $y$.
$\sin(x)$ is concave on $[0,\pi]$. Therefore,
$$
\frac{\sin(x)}{x}=\frac{\sin(x)-\sin(0)}{x-0}
$$
is a decreasing function of $x$. Therefore, if $\beta\lt\alpha$, then
$$
\frac{\sin(\beta)}{\beta}\gt\frac{\sin(\alpha)}{\alpha}\implies\frac{\alpha}{\beta}\gt\frac{\sin(\alpha)}{\sin(\beta)}
$$

Sine Doubling and $\boldsymbol{\lim\limits_{x\to0}\frac{\sin(x)}x=1}$
Using $\cos(x/2)=\frac{\sin(x)}{2\sin(x/2)}$, induction gives
$$
\prod_{k=1}^n\cos\left(\frac{x}{2^k}\right)=\frac{\sin(x)}{2^n\sin\left(\frac{x}{2^n}\right)}
$$
As shown in this answer, $\lim\limits_{x\to0}\frac{\sin(x)}x=1$. Therefore, $\lim\limits_{n\to\infty}2^n\sin\left(\frac{x}{2^n}\right)=x$. Thus,
$$
\prod_{k=1}^\infty\cos\left(\frac{x}{2^k}\right)=\frac{\sin(x)}x
$$
Since $\cos(x)$ is decreasing on $\left[0,\frac\pi2\right]$, we see that $\frac{\sin(x)}x$ is decreasing on $[0,\pi]$. Therefore, if $\beta\lt\alpha$, then
$$
\frac{\sin(\beta)}{\beta}\gt\frac{\sin(\alpha)}{\alpha}\implies\frac{\alpha}{\beta}\gt\frac{\sin(\alpha)}{\sin(\beta)}
$$
A: $$f(x)={x \over \sin x}$$
$$f'(x)={\sin x - x \cos x \over \sin^2x}={\tan x - x \over \sin^2x \cos x}$$
For $x\in(0, \pi/2)$, $\tan x>x$ (proof) so $f'(x)>0$ which means that the function $f(x)$ is monotoniously increasing.
So if $\alpha>\beta$:
$$f(\alpha)>f(\beta)$$
$${\alpha \over \sin \alpha}>{\beta \over \sin \beta}$$
$${\alpha \over \beta}>{\sin \alpha \over \sin \beta}$$
