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 α and β are acute angles (i.e. between 0 and a right angle) and β < α then

[math]\displaystyle{ \frac{\sin\alpha}{\sin\beta} \lt \frac{\alpha}{\beta} \lt \frac{\tan\alpha}{\tan\beta}. }[/math]

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

[math]\displaystyle{ 0\lt \sin(\alpha)\lt \alpha\lt \tan(\alpha) }[/math],

[math]\displaystyle{ 0\lt \sin(\beta)\lt \sin(\alpha)\lt 1 }[/math] and

[math]\displaystyle{ 1\gt \cos(\beta)\gt \cos(\alpha)\gt 0 }[/math].

Proof of the first inequality

Using these inequalities we can first prove that

[math]\displaystyle{ \frac{\sin(\alpha)}{\sin(\beta)} \lt \frac{\alpha}{\beta}. }[/math]

We first note that the inequality is equivalent to

[math]\displaystyle{ \frac{\sin(\alpha)}{\alpha} \lt \frac{\sin(\beta)}{\beta} }[/math]

which itself can be rewritten as

[math]\displaystyle{ \frac{\sin(\alpha)-\sin(\beta)}{\alpha-\beta} \lt \frac{\sin(\beta)}{\beta}. }[/math]

We now want show that

[math]\displaystyle{ \frac{\sin(\alpha)-\sin(\beta)}{\alpha-\beta}\lt \cos(\beta) \lt \frac{\sin(\beta)}{\beta}. }[/math]

The second inequality is simply [math]\displaystyle{ \beta\lt \tan\beta }[/math]. The first one is true because

[math]\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} \lt \frac{2\cdot \left(\frac{\alpha-\beta}2 \right) \cdot \cos(\beta)}{\alpha-\beta} = \cos(\beta). }[/math]

Proof of the second inequality

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

[math]\displaystyle{ \frac{\alpha}{\beta} \lt \frac{\tan(\alpha)}{\tan(\beta)}. }[/math]

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

[math]\displaystyle{ \beta\lt \tan(\beta)=\frac{\sin(\beta)}{\cos(\beta)}\lt \frac{\sin(\beta)}{\cos(\alpha)} }[/math]

Consequently, using that [math]\displaystyle{ 0\lt \alpha-\beta\lt \alpha }[/math] in the previous equation (replacing [math]\displaystyle{ \beta }[/math] by [math]\displaystyle{ \alpha-\beta\lt \alpha }[/math]) we obtain:

[math]\displaystyle{ {\alpha-\beta}\lt {\frac{\sin(\alpha-\beta)}{\cos(\alpha)}}=\tan(\alpha)\cos(\beta)-\sin(\beta). }[/math]

We conclude that

[math]\displaystyle{ \frac{\alpha}{\beta}=\frac{\alpha-\beta}{\beta}+1\lt \frac{\tan(\alpha)\cos(\beta)-\sin(\beta)}{\sin(\beta)}+1 = \frac{\tan(\alpha)}{\tan(\beta)}. }[/math]

See also

• Aristarchus of Samos
• Eratosthenes
• Posidonius

External links

• Leibowitz, Gerald M.. "Hellenistic Astronomers and the Origins of Trigonometry". https://www2.math.uconn.edu/~leibowitz/math2720s11/Greek_Trig.pdf. 
• Proof of the First Inequality
• Proof of the Second Inequality