Geometrically showing $\frac{\alpha}{\beta} > \frac{\sin\alpha}{\sin\beta}$, for $0 < \beta < \alpha < 90^\circ$

Question

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 $\frac{\sin\alpha}{\sin\beta}$?

Answer 1

$$f(x)={x \over \sin x}$$

$$f'(x)={\sin x - x \cos x \over \sin^2x}={\tan x - x \over \sin^2x \cos x}$$

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 $\alpha>\beta$:

$$f(\alpha)>f(\beta)$$

$${\alpha \over \sin \alpha}>{\beta \over \sin \beta}$$

$${\alpha \over \beta}>{\sin \alpha \over \sin \beta}$$

Answer 2

$$\sin x < x < \tan x \qquad \forall~x \in [0, \frac{\pi}2]$$

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. $\frac{ \sin\beta }{ \cos( \frac{ \alpha + \beta}2 ) }$).

Answer 3

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 $$ \frac{\sin(\beta)}{\beta}\gt\frac{\sin(\alpha)}{\alpha}\implies\frac{\alpha}{\beta}\gt\frac{\sin(\alpha)}{\sin(\beta)} $$

Answer 4

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.)