Solution:

\begin{aligned} \cos 20^{0} \cos 40^{0} \cos 60^{0} \cos 80^{0} &=\frac{1}{2}\left(2 \cos 20^{\circ} \cos 40^{\circ}\right) \cdot\left(\frac{1}{2}\right) \cos 80^{\circ} \ &=\frac{1}{4}\left[\cos \left(20^{\circ}+40^{\circ}\right)+\cos \left(20^{\circ}-40^{\circ}\right)\right] \cos 80^{\circ} \ & =\frac{1}{4}\left[\cos \left(60^{\circ}\right)+\cos \left(-20^{\circ}\right)\right] \cos 80^{\circ}\ & =\frac{1}{4}\left[\frac{1}{2} \cos 80^{\circ}+\cos 20^{\circ} \cos 80^{\circ}\right] \ & =\frac{1}{4}\left[\frac{1}{2} \cos 80^{\circ}+\frac{1}{2}\left(2 \cos 20^{\circ} \cos 80^{\circ}\right)\right]\ & =\frac{1}{8}\left[\cos 80^{\circ}+\cos …