Aristarchus's inequality explained

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.

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

\sin(\alpha)
\sin(\beta)

<

\alpha
\beta

.

We first note that the inequality is equivalent to

\sin(\alpha)
\alpha

<

\sin(\beta)
\beta

which itself can be rewritten as
\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

. The first one is true because
\sin(\alpha)-\sin(\beta)
\alpha-\beta

=

2 ⋅ \sin\left(\alpha-\beta
2 \right)\cos\left(\alpha+\beta
2\right)
\alpha-\beta

<

2 ⋅
\left(\alpha-\beta
2 \right)
\cos(\beta)
\alpha-\beta

=\cos(\beta).

Proof of the second inequality

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

in the previous equation (replacing

\beta

by

\alpha-\beta<\alpha

) we obtain:
{\alpha-\beta}<{\sin(\alpha-\beta)
\cos(\alpha)
}=\tan(\alpha)\cos(\beta)-\sin(\beta).We conclude that
\alpha=
\beta
\alpha-\beta
\beta

+1<

\tan(\alpha)\cos(\beta)-\sin(\beta)
\sin(\beta)

+1=

\tan(\alpha)
\tan(\beta)

.

See also

External links