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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
![{\displaystyle {\frac {\sin \alpha }{\sin \beta }}<{\frac {\alpha }{\beta }}<{\frac {\tan \alpha }{\tan \beta }}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a3d279d7de64b2f6768c21857bb04b5caa225788)
Ptolemy used the first of these inequalities while constructing his table of chords.[1]
The proof is a consequence of the more widely known inequalities
,
and
.
Proof of the first inequality
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Using these inequalities we can first prove that
![{\displaystyle {\frac {\sin(\alpha )}{\sin(\beta )}}<{\frac {\alpha }{\beta }}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c0251d931871412c75038e56f889218542ca62e1)
We first note that the inequality is equivalent to
![{\displaystyle {\frac {\sin(\alpha )}{\alpha }}<{\frac {\sin(\beta )}{\beta }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/507349d683a8eeac6f576722c9cc5bd5b95c4709)
which itself can be rewritten as
![{\displaystyle {\frac {\sin(\alpha )-\sin(\beta )}{\alpha -\beta }}<{\frac {\sin(\beta )}{\beta }}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/24834ac9af8076511c195e1d9f861e7bcb99c79d)
We now want show that
![{\displaystyle {\frac {\sin(\alpha )-\sin(\beta )}{\alpha -\beta }}<\cos(\beta )<{\frac {\sin(\beta )}{\beta }}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8c2884c4002e4bbeba86273653697af4743f66f6)
The second inequality is simply
. 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 ).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/340badef325cfdb5eb9c798dac1e88be89bebf01)
Proof of the second inequality
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Now we want to show the second inequality, i.e. that:
![{\displaystyle {\frac {\alpha }{\beta }}<{\frac {\tan(\alpha )}{\tan(\beta )}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/007de3ba152a45c32fe1bb77eddba4cda97537e6)
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 )}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/66d65db7d7802ead7f7d8157303e38a7d39756cb)
Consequently, using that
in the previous equation (replacing
by
) we obtain:
![{\displaystyle {\alpha -\beta }<{\frac {\sin(\alpha -\beta )}{\cos(\alpha )}}=\tan(\alpha )\cos(\beta )-\sin(\beta ).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/22f4153c5f6b00c057d578f6e45a450881356c7e)
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 )}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c1a370b15621d334a4c98ce60f3418ccbbbbe6cd)
Notes and references
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