Question

Вычислите значение выражения log2sinπ/12 + log2sinπ/6 + log2sin5π/12.

Answer 1

$log_2sin\frac{\pi}{12}+log_2sin\frac{\pi}{6}+log_2sin\frac{5\pi}{12}=log_2sin\frac{\pi}{12}*sin(\frac{\pi}{2}-\frac{\pi}{12})+$$log_2sin\frac{\pi}{6}=log_2sin\frac{\pi}{12}cos\frac{\pi}{12}+log_2sin\frac{\pi}{6}=log_2\frac{sin\frac{\pi}{6}}{2}+log_2sin\frac{\pi}{6}=$$log_2\frac{sin\frac{\pi}{6}}{2}+log_2sin\frac{\pi}{6}=log_2\frac{1}{4}+log_2\frac{1}{2}=-2-1=-3$