Log2cos20+log2cos40+log2cos80

Решите задачу:

$log_2(cos20)+log_2(cos40)+log_2(cos80)=\\\\=log_2(cos20\cdot cos40\cdot cos80)=log_2\frac{1}{8}=log_22^{-3}=-3\\\\\\cos20\cos40\cdot cos80= \frac{(sin20\cdot cos20)\cdot cos40\cdot cos80}{sin20} =\\\\= \frac{(\frac{1}{2}sin40\cdot cos40)\cdot cos80}{sin20} = \frac{\frac{1}{2}\cdot \frac{1}{2}sin80 \cdot cos80}{sin20}= \frac{\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot sin160}{sin20} =\\\\=\frac{1}{8}\cdot \frac{sin(180-20)}{sin20} =\frac{1}{8}\cdot \frac{sin20}{sin20} =\frac{1}{8}=2^{-3}$