3
$\begingroup$

enter image description here

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

$\endgroup$
5
  • 1
    $\begingroup$ Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps. $\endgroup$ – robjohn Nov 1 '18 at 15:30
  • $\begingroup$ @robjohn I have identified the line segments equal to one and a few that are equal to $\sin\alpha$ and $\sin\beta$ (I now see this is difficult to explain without points on my diagram). My question is how to get the two sin over each other? I'm aware that tan is sin/cos but do not know how to approach sin/sin? Maybe a ratio? $\endgroup$ – K Math Nov 1 '18 at 15:35
  • 2
    $\begingroup$ As mentioned in the post on context I linked above, where this problem came from can be important. This is not a simple problem and the simplest approach I see so far is to use the identity $$\frac{\sin(x)}{x}=\prod_{k=1}^\infty\cos\left(\frac x{2^k}\right)$$ and then notice that $\cos(x)$ is a decreasing function on $\left[0,\frac\pi2\right]$. $\endgroup$ – robjohn Nov 1 '18 at 15:41
  • 1
    $\begingroup$ Also note that the inequality, as stated in the question, cannot be correct since taking the reciprocal of both sides gives $\frac\beta\alpha\lt\frac{\sin(\beta)}{\sin(\alpha)}$. You need the assumption shown in the image that $\beta\lt\alpha$. $\endgroup$ – robjohn Nov 1 '18 at 15:55
  • 1
    $\begingroup$ Although I had come across it before, it had entirely slipped my mind that this inequality is associated with the name of Aristarchus of Samos (c.310-230 BC). A search including his name turns up this answer by @EuYu: Proof of Aristarchus' Inequality. $\endgroup$ – Calum Gilhooley Nov 13 '18 at 23:44
3
$\begingroup$

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 $n = 5$, $m = 3$, $j = 2$:

An illustration of 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.)

$\endgroup$
3
  • 1
    $\begingroup$ @TangentPlane I'm getting a little frazzled, so will have to take some time to digest all this. Meanwhile, I have added a section about perimeters to my question in HSM-SE. I apologise if it is essentially just a copy of something you've already written! $\endgroup$ – Calum Gilhooley Nov 8 '18 at 20:37
  • 1
    $\begingroup$ Alternatively, since the angles $\angle A_j A_0 A_{j+1}$ are all equal for $1\leq j < n,$ you can reflect $A_{j-1}$ over $A_0A_j$ onto $A_0 A_{j+1}$, and then $\triangle A_{j-1} A_0 A_j$ is congruent to a triangle that is strictly contained in $\triangle A_j A_0 A_{j+1}$. $\endgroup$ – David K Feb 10 '20 at 13:58
  • $\begingroup$ @DavidK That's a lot neater. Thank you. (Incidentally, it looks as if a long comment thread has been deleted. I never did get to grips with all the implications of this question, or this, or several other related questions. A big tidying-up job is probably needed.) $\endgroup$ – Calum Gilhooley Feb 10 '20 at 16:02
3
$\begingroup$

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

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

$\endgroup$
0
$\begingroup$

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

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

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 $\alpha>\beta$:

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

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

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.