Prove that, cos20° cos40° cos80°=$\frac{1}{8}$
Solution:
cos20° cos40° cos80°
= $\frac{1}{2}$ (2 cos20° cos40°) cos80°
= $\frac{1}{2}$ {cos (20°+40°) + cos(20°-40°)} cos80°
= $\frac{1}{2}$ {cos60° +cos(-20°)} cos80°
= $\frac{1}{2}$ (cos60° +cos20°) cos80°
=$\frac{1}{2}$ ($\frac{1}{2}$ +cos20°) cos80°
= $\frac{1}{4}$ cos80° +$\frac{1}{2}$ cos20° cos80°
= $\frac{1}{4}$ cos80°+ $\frac{1}{4}$ (2cos20° cos80°)
= $\frac{1}{4}$ cos80°+ $\frac{1}{4}$ {cos(20°+80°) + cos(20°-80°)}
= $\frac{1}{4}$ cos80°+ $\frac{1}{4}$ {cos100° +cos(-60°)}
= $\frac{1}{4}$ cos80°+$\frac{1}{4}$ (cos100° +cos60°)
= $\frac{1}{4}$ cos80°+ $\frac{1}{4}$ cos(180°-80°) +$\frac{1}{4}$ cos60°
= $\frac{1}{4}$ cos80°+$\frac{1}{4}$(-cos80°) + $\frac{1}{4}$ ×$\frac{1}{2}$
= $\frac{1}{4}$ cos80° – $\frac{1}{4}$ cos80°+$\frac{1}{8}$
= $\frac{1}{8}$ [Proved]