MHB Trigonometry challenge - cosine product

[Greg]

Prove $$\cos20^\circ\cdot\cos40^\circ\cdot\cos80^\circ=\frac18$$

 

[anemone]

Prove $$\cos20^\circ\cdot\cos40^\circ\cdot\cos80^\circ=\frac18$$

My solution:

Spoiler

$\begin{align*}\cos 60^{\circ}&=4\cos^3 20^{\circ}-3\cos 20^{\circ}\\&=\cos 20^{\circ}(4\cos^2 20^{\circ}-3)\\&=\cos 20^{\circ}(2(2\cos^2 20^{\circ}-1)-1)\\&=\cos 20^{\circ}(2(\cos 40^{\circ})-1)\\&=\cos 20^{\circ}(2\cos 40^{\circ}+2\cos 120^{\circ})\\&=2\cos 20^{\circ}(\cos 40^{\circ}+\cos 120^{\circ})\\&=2\cos 20^{\circ}(2\cos 40^{\circ}\cos 80^{\circ})\\&=4\cos 20^{\circ}\cos 40^{\circ}\cos 80^{\circ}\end{align*}$

$\therefore \cos 20^{\circ}\cos 40^{\circ}\cos 80^{\circ}=\dfrac{1}{8}$

 

[kaliprasad]

Prove $$\cos20^\circ\cdot\cos40^\circ\cdot\cos80^\circ=\frac18$$

Spoiler

$8 \sin 20^\circ\cdot \cos20^\circ\cdot\cos40^\circ\cdot\cos80^\circ$
= $4 \sin 40^\circ \cdot\cos40^\circ\cdot\cos80^\circ$
= $2 \sin 80^\circ \cdot \cos80^\circ$
= $\sin 160^\circ$
= $\sin 20^\circ$
dividing both sides by $8 \sin 20^\circ$ we get the result