Trigonometric exercise asked by Dr. Yousef abbas in theمنبر الرياضيات facebook group

Exercise:

Evaluate the following, $\cos \frac{\pi }{7}\cos \frac{2\pi }{7}%
\cos \frac{4\pi }{7}$

Solution:

we have A=$\cos \theta \cos 2\theta \cos 4\theta $ then

$\sin \theta A=\sin \theta \cos \theta \cos 2\theta \cos 4\theta $


$2\sin \theta A=\sin 2\theta \cos 2\theta \cos 4\theta $

$4\sin \theta A=\sin 4\theta \cos 4\theta $

$8\sin \theta A=\sin 8\theta $

thus, $A=\frac{\sin 8\theta }{8\sin \theta }$

but $\sin 8\theta =\sin \left( \frac{8\pi }{7}\right) =\sin \left( \pi +%
\frac{\pi }{7}\right) =-\sin \left( \frac{\pi }{7}\right) =-\sin \theta $

Therefore, $A=\frac{-\sin \theta }{8\sin \theta }=-\frac{1}{8}$