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## Summary

Aristarchus's inequality (after the Greek astronomer and mathematician Aristarchus of Samos; c. 310 – c. 230 BCE) is a law of trigonometry which states that if α and β are acute angles (i.e. between 0 and a right angle) and β < α then

${\frac {\sin \alpha }{\sin \beta }}<{\frac {\alpha }{\beta }}<{\frac {\tan \alpha }{\tan \beta }}.$ Ptolemy used the first of these inequalities while constructing his table of chords.

## Proof

The proof is a consequence of the more known inequalities

$0<\sin(\alpha )<\alpha <\tan(\alpha )$ ,
$0<\sin(\beta )<\sin(\alpha )<1$  and
$1>\cos(\beta )>\cos(\alpha )>0$ .

### Proof of the first inequality

Using these inequalities we can first prove that

${\frac {\sin(\alpha )}{\sin(\beta )}}<{\frac {\alpha }{\beta }}.$

We first note that the inequality is equivalent to

${\frac {\sin(\alpha )}{\alpha }}<{\frac {\sin(\beta )}{\beta }}$

which itself can be rewritten as

${\frac {\sin(\alpha )-\sin(\beta )}{\alpha -\beta }}<{\frac {\sin(\beta )}{\beta }}.$

We now want show that

${\frac {\sin(\alpha )-\sin(\beta )}{\alpha -\beta }}<\cos(\beta )<{\frac {\sin(\beta )}{\beta }}.$

The second inequality is simply $\beta <\tan \beta$ . The first one is true because

${\frac {\sin(\alpha )-\sin(\beta )}{\alpha -\beta }}={\frac {2\cdot \sin \left({\frac {\alpha -\beta }{2}}\right)\cos \left({\frac {\alpha +\beta }{2}}\right)}{\alpha -\beta }}<{\frac {2\cdot \left({\frac {\alpha -\beta }{2}}\right)\cdot \cos(\beta )}{\alpha -\beta }}=\cos(\beta ).$

### Proof of the second inequality

Now we want to show the second inequality, i.e. that:

${\frac {\alpha }{\beta }}<{\frac {\tan(\alpha )}{\tan(\beta )}}.$

We first note that due to the initial inequalities we have that:

$\beta <\tan(\beta )={\frac {\sin(\beta )}{\cos(\beta )}}<{\frac {\sin(\beta )}{\cos(\alpha )}}$

Consequently, using that $0<\alpha -\beta <\alpha$  in the previous equation (replacing $\beta$  by $\alpha -\beta <\alpha$ ) we obtain:

${\alpha -\beta }<{\frac {\sin(\alpha -\beta )}{\cos(\alpha )}}=\tan(\alpha )\cos(\beta )-\sin(\beta ).$

We conclude that

${\frac {\alpha }{\beta }}={\frac {\alpha -\beta }{\beta }}+1<{\frac {\tan(\alpha )\cos(\beta )-\sin(\beta )}{\sin(\beta )}}+1={\frac {\tan(\alpha )}{\tan(\beta )}}.$