Harvard-MIT Math Tournament

Solution:

We will use the identities $\cos a + \cos b = 2 \cos \frac{a+b}{2} \cos \frac{a-b}{2}$ and $\sin a + \sin b = 2 \sin \frac{a+b}{2} \cos \frac{a-b}{2}$.

The numerator is
$$(\sin 10 + \sin 80) + (\sin 20 + \sin 70) + (\sin 30 + \sin 60) + (\sin 40 + \sin 50)$$

$= 2 \sin 45 (\cos 35 + \cos 25 + \cos 15 + \cos 5)$

$= 2 \sin 45 ((\cos 35 + \cos 5) + (\cos 25 + \cos 15))$

$= 4 \sin 45 \cos 20 (\cos 15 + \cos 5)$

$= 8 \sin 45 \cos 20 \cos 10 \cos 5$

So the fraction equals $8 \sin 45 = 4 \sqrt{2}$.