Question

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

Answer 1

$log_2sinfrac{pi}{12}+log_2sinfrac{pi}{6}+log_2sinfrac{5pi}{12}=log_2sinfrac{pi}{12}*sin(frac{pi}{2}-frac{pi}{12})+$$log_2sinfrac{pi}{6}=log_2sinfrac{pi}{12}cosfrac{pi}{12}+log_2sinfrac{pi}{6}=log_2frac{sinfrac{pi}{6}}{2}+log_2sinfrac{pi}{6}=$$log_2frac{sinfrac{pi}{6}}{2}+log_2sinfrac{pi}{6}=log_2frac{1}{4}+log_2frac{1}{2}=-2-1=-3$