Aristarchus's inequality

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

sin ⁡ α sin ⁡ β < α β < tan ⁡ α tan ⁡ β . {\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

0 < sin ⁡ ( α ) < α < tan ⁡ ( α ) {\displaystyle 0<\sin(\alpha )<\alpha <\tan(\alpha )} ,
0 < sin ⁡ ( β ) < sin ⁡ ( α ) < 1 {\displaystyle 0<\sin(\beta )<\sin(\alpha )<1} and
1 > cos ⁡ ( β ) > cos ⁡ ( α ) > 0 {\displaystyle 1>\cos(\beta )>\cos(\alpha )>0} .

Proof of the first inequality

Using these inequalities we can first prove that

sin ⁡ ( α ) sin ⁡ ( β ) < α β . {\displaystyle {\frac {\sin(\alpha )}{\sin(\beta )}}<{\frac {\alpha }{\beta }}.}

We first note that the inequality is equivalent to

sin ⁡ ( α ) α < sin ⁡ ( β ) β {\displaystyle {\frac {\sin(\alpha )}{\alpha }}<{\frac {\sin(\beta )}{\beta }}}

which itself can be rewritten as

sin ⁡ ( α ) − sin ⁡ ( β ) α − β < sin ⁡ ( β ) β . {\displaystyle {\frac {\sin(\alpha )-\sin(\beta )}{\alpha -\beta }}<{\frac {\sin(\beta )}{\beta }}.}

We now want show that

sin ⁡ ( α ) − sin ⁡ ( β ) α − β < cos ⁡ ( β ) < sin ⁡ ( β ) β . {\displaystyle {\frac {\sin(\alpha )-\sin(\beta )}{\alpha -\beta }}<\cos(\beta )<{\frac {\sin(\beta )}{\beta }}.}

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

sin ⁡ ( α ) − sin ⁡ ( β ) α − β = 2 ⋅ sin ⁡ ( α − β 2 ) cos ⁡ ( α + β 2 ) α − β < 2 ⋅ ( α − β 2 ) ⋅ cos ⁡ ( β ) α − β = cos ⁡ ( β ) . {\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:

α β < tan ⁡ ( α ) tan ⁡ ( β ) . {\displaystyle {\frac {\alpha }{\beta }}<{\frac {\tan(\alpha )}{\tan(\beta )}}.}

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

β < tan ⁡ ( β ) = sin ⁡ ( β ) cos ⁡ ( β ) < sin ⁡ ( β ) cos ⁡ ( α ) {\displaystyle \beta <\tan(\beta )={\frac {\sin(\beta )}{\cos(\beta )}}<{\frac {\sin(\beta )}{\cos(\alpha )}}}

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

α − β < sin ⁡ ( α − β ) cos ⁡ ( α ) = tan ⁡ ( α ) cos ⁡ ( β ) − sin ⁡ ( β ) . {\displaystyle {\alpha -\beta }<{\frac {\sin(\alpha -\beta )}{\cos(\alpha )}}=\tan(\alpha )\cos(\beta )-\sin(\beta ).}

We conclude that

α β = α − β β + 1 < tan ⁡ ( α ) cos ⁡ ( β ) − sin ⁡ ( β ) sin ⁡ ( β ) + 1 = tan ⁡ ( α ) tan ⁡ ( β ) . {\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

  1. ^ Toomer, G. J. (1998), Ptolemy's Almagest, Princeton University Press, p. 54, ISBN 0-691-00260-6

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
  • v
  • t
  • e
Ancient Greek mathematics
Mathematicians
(timeline)Treatises
ProblemsConcepts
and definitionsResults
In Elements
Apollonius
Other
CentersRelated
History of
Other cultures
 Ancient Greece portal • icon Mathematics portal
Stub icon

This elementary geometry-related article is a stub. You can help Wikipedia by expanding it.

  • v
  • t
  • e