Question

log2cos20+log2cos40+log2cos80

Answer 1

$log_2(cos20)+log_2(cos40)+log_2(cos80)=\\=log_2(cos20cdot cos40cdot cos80)=log_2frac{1}{8}=log_22^{-3}=-3\\\cos20cos40cdot cos80= frac{(sin20cdot cos20)cdot cos40cdot cos80}{sin20} =\\= frac{(frac{1}{2}sin40cdot cos40)cdot cos80}{sin20} = frac{frac{1}{2}cdot frac{1}{2}sin80 cdot cos80}{sin20}= frac{frac{1}{2}cdot frac{1}{2}cdot frac{1}{2}cdot sin160}{sin20} =\\=frac{1}{8}cdot frac{sin(180-20)}{sin20} =frac{1}{8}cdot frac{sin20}{sin20} =frac{1}{8}=2^{-3}$