# Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 6"

## Problem

If $\tan 15^\circ \tan 25^\circ \tan 35^\circ =\tan \theta$ and $0^\circ \le \theta \le 180^\circ,$ find $\theta.$

## Solution

We know from product to sum formulas we have: $$\frac{\sin 15^\circ\sin 25^\circ\sin 35^\circ}{\cos 15^\circ\cos 25^\circ\cos 35^\circ}=\frac{\sin 15^{\circ}(\cos 10^\circ-\cos 60^\circ)}{\cos 15^\circ(\cos 10^\circ+\cos 60^\circ)}$$ Multiply by $\frac{2}{2}$: $$\frac{2\sin 15^\circ\cos 10^\circ-\sin 15^\circ}{\cos 15^\circ+2\cos 15^\circ\cos 10^\circ}$$ Again use product to sum: $$\frac{\sin 5^\circ-\sin 15^\circ+\sin 25^\circ}{\cos 5^\circ+\cos 15^\circ+\cos 25^\circ}$$ Finally, use sum to product on the rightmost terms in the numerator and denominator: $$\frac{\sin 5^{\circ}+2\sin 5^\circ\cos 20^\circ}{\cos 5^\circ+2\cos 5^\circ\cos 20^\circ}=\frac{\sin 5^\circ(1+2\cos 20^\circ)}{\cos 5^\circ(1+2\cos 20^\circ)}=\tan 5^\circ$$ Thus, $\theta=\boxed{005}$.