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 &alpha; and &beta; are acute angles (i.e. between 0 and a right angle) and &beta; < &alpha; then


 * $$ \frac{\sin\alpha}{\sin\beta} < \frac{\alpha}{\beta} < \frac{\tan\alpha}{\tan\beta}. $$

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


 * $$ 0<\sin(\alpha)<\alpha<\tan(\alpha) $$,


 * $$ 0<\sin(\beta)<\sin(\alpha)<1 $$ and


 * $$ 1>\cos(\beta)>\cos(\alpha)>0$$.

Proof of the first inequality
Using these inequalities we can first prove that


 * $$ \frac{\sin(\alpha)}{\sin(\beta)} < \frac{\alpha}{\beta}. $$

We first note that the inequality is equivalent to
 * $$\frac{\sin(\alpha)}{\alpha} < \frac{\sin(\beta)}{\beta} $$

which itself can be rewritten as
 * $$\frac{\sin(\alpha)-\sin(\beta)}{\alpha-\beta} < \frac{\sin(\beta)}{\beta}. $$

We now want show that
 * $$\frac{\sin(\alpha)-\sin(\beta)}{\alpha-\beta}<\cos(\beta) < \frac{\sin(\beta)}{\beta}. $$

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

\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:


 * $$ \frac{\alpha}{\beta} <\frac{\tan(\alpha)}{\tan(\beta)}. $$

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


 * $$ \beta<\tan(\beta)=\frac{\sin(\beta)}{\cos(\beta)}<\frac{\sin(\beta)}{\cos(\alpha)} $$

Consequently, using that $$0<\alpha-\beta<\alpha $$ in the previous equation (replacing $$\beta $$ by $$\alpha-\beta<\alpha $$) we obtain:
 * $${\alpha-\beta}<{\frac{\sin(\alpha-\beta)}{\cos(\alpha)}}=\tan(\alpha)\cos(\beta)-\sin(\beta).$$

We conclude that
 * $$ \frac{\alpha}{\beta}=\frac{\alpha-\beta}{\beta}+1< \frac{\tan(\alpha)\cos(\beta)-\sin(\beta)}{\sin(\beta)}+1 = \frac{\tan(\alpha)}{\tan(\beta)}. $$