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.
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 to 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
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