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\alpha | |
\sin\beta |
<
\alpha | |
\beta |
<
\tan\alpha | |
\tan\beta |
.
Ptolemy used the first of these inequalities while constructing his table of chords.
The proof is a consequence of the more widely known inequalities
0<\sin(\alpha)<\alpha<\tan(\alpha)
0<\sin(\beta)<\sin(\alpha)<1
1>\cos(\beta)>\cos(\alpha)>0
Using these inequalities we can first prove that
\sin(\alpha) | |
\sin(\beta) |
<
\alpha | |
\beta |
.
We first note that the inequality is equivalent to
\sin(\alpha) | |
\alpha |
<
\sin(\beta) | |
\beta |
\sin(\alpha)-\sin(\beta) | |
\alpha-\beta |
<
\sin(\beta) | |
\beta |
.
We now want show that
\sin(\alpha)-\sin(\beta) | |
\alpha-\beta |
<\cos(\beta)<
\sin(\beta) | |
\beta |
.
The second inequality is simply
\beta<\tan\beta
\sin(\alpha)-\sin(\beta) | |
\alpha-\beta |
=
| |||||||
\alpha-\beta |
<
| ||||||
\alpha-\beta |
=\cos(\beta).
Now we want to show the second inequality, i.e. that:
\alpha | < | |
\beta |
\tan(\alpha) | |
\tan(\beta) |
.
We first note that due to the initial inequalities we have that:
\beta<\tan(\beta)= | \sin(\beta) | < |
\cos(\beta) |
\sin(\beta) | |
\cos(\alpha) |
Consequently, using that
0<\alpha-\beta<\alpha
\beta
\alpha-\beta<\alpha
{\alpha-\beta}<{ | \sin(\alpha-\beta) |
\cos(\alpha) |
\alpha | = | |
\beta |
\alpha-\beta | |
\beta |
+1<
\tan(\alpha)\cos(\beta)-\sin(\beta) | |
\sin(\beta) |
+1=
\tan(\alpha) | |
\tan(\beta) |
.