Найдите значение выражения log2sin(5π/12) + log2cos(5π/12)
log2sin(5π/12)+log2cos(5π/12)= =log2(sin(5π/12)·cos(5π/12))= =log2(1/2)·sin(5π/6)= =log2(1/2)+log2sin(5π/6) =log21/2+log2(1/2)=–1–1=–2