Aristarchus's inequality

Aristarchus's inequality (after the Greek

(i.e. between 0 and a right angle) and β < α then

${\displaystyle {\frac {\sin \alpha }{\sin \beta }}<{\frac {\alpha }{\beta }}<{\frac {\tan \alpha }{\tan \beta }}.}$

Ptolemy used the first of these inequalities while constructing his table of chords.^[1]

Proof

The proof is a consequence of the more widely known inequalities

${\displaystyle 0<\sin(\alpha )<\alpha <\tan(\alpha )}$,

${\displaystyle 0<\sin(\beta )<\sin(\alpha )<1}$ and

${\displaystyle 1>\cos(\beta )>\cos(\alpha )>0}$.

Proof of the first inequality

Using these inequalities we can first prove that

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

We first note that the inequality is equivalent to

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

which itself can be rewritten as

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

We now want show that

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

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

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

Proof of the second inequality

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

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

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

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

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

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

We conclude that

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

See also

• Aristarchus of Samos
• Eratosthenes
• Posidonius

Notes and references

External links

• Leibowitz, Gerald M. "Hellenistic Astronomers and the Origins of Trigonometry" (PDF). Archived from the original (PDF) on 2011-09-27. Retrieved 2019-06-24.
• Proof of the First Inequality
• Proof of the Second Inequality

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