Aristarchus's inequality

Template:One source 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

\frac{\sin α}{\sin β} < \frac{α}{β} < \frac{\tan α}{\tan β} .

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)

,

0 < \sin (β) < \sin (α) < 1

and

1 > \cos (β) > \cos (α) > 0

.

Using these inequalities we can first prove that

\frac{\sin (α)}{\sin (β)} < \frac{α}{β} .

We first note that the inequality is equivalent to

\frac{\sin (α)}{α} < \frac{\sin (β)}{β}

which itself can be rewritten as

\frac{\sin (α) - \sin (β)}{α - β} < \frac{\sin (β)}{β} .

We now want to show that

\frac{\sin (α) - \sin (β)}{α - β} < \cos (β) < \frac{\sin (β)}{β} .

The second inequality is simply $β < \tan β$ . The first one is true because

\frac{\sin (α) - \sin (β)}{α - β} = \frac{2 \cdot \sin (\frac{α - β}{2}) \cos (\frac{α + β}{2})}{α - β} < \frac{2 \cdot (\frac{α - β}{2}) \cdot \cos (β)}{α - β} = \cos (β) .

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

\frac{α}{β} < \frac{\tan (α)}{\tan (β)} .

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

β < \tan (β) = \frac{\sin (β)}{\cos (β)} < \frac{\sin (β)}{\cos (α)}

Consequently, using that $0 < α - β < α$ in the previous equation (replacing $β$ by $α - β < α$ ) we obtain:

α - β < \frac{\sin (α - β)}{\cos (α)} = \tan (α) \cos (β) - \sin (β) .

We conclude that

\frac{α}{β} = \frac{α - β}{β} + 1 < \frac{\tan (α) \cos (β) - \sin (β)}{\sin (β)} + 1 = \frac{\tan (α)}{\tan (β)} .