Aristarchus's inequality

Using these inequalities we can first prove that

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

We first note that the inequality is equivalent to ${\displaystyle {\frac {\sin(\alpha )}{\alpha }}<{\frac {\sin(\beta )}{\beta }}}$ which itself can be rewritten as ${\displaystyle {\frac {\sin(\alpha )-\sin(\beta )}{\alpha -\beta }}<{\frac {\sin(\beta )}{\beta }}.}$

We now want show that

${\displaystyle {\frac {\sin(\alpha )-\sin(\beta )}{\alpha -\beta }}<\cos(\beta )<{\frac {\sin(\beta )}{\beta }}.}$

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

${\displaystyle {\frac {\sin(\alpha )-\sin(\beta )}{\alpha -\beta }}={\frac {2\cdot \sin \left({\frac {\alpha -\beta }{2}}\right)\cos \left({\frac {\alpha +\beta }{2}}\right)}{\alpha -\beta }}<{\frac {2\cdot \left({\frac {\alpha -\beta }{2}}\right)\cdot \cos(\beta )}{\alpha -\beta }}=\cos(\beta ).}$

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

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

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

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

Consequently, using that ${\displaystyle 0<\alpha -\beta <\alpha }$ in the previous equation (replacing ${\displaystyle \beta }$ by ${\displaystyle \alpha -\beta <\alpha }$) we obtain:

${\displaystyle {\alpha -\beta }<{\frac {\sin(\alpha -\beta )}{\cos(\alpha )}}=\tan(\alpha )\cos(\beta )-\sin(\beta ).}$

We conclude that

${\displaystyle {\frac {\alpha }{\beta }}={\frac {\alpha -\beta }{\beta }}+1<{\frac {\tan(\alpha )\cos(\beta )-\sin(\beta )}{\sin(\beta )}}+1={\frac {\tan(\alpha )}{\tan(\beta )}}.}$