Найдите значение выражения log2sin(5Pi/12) + log2cos(5Pi/12)
log_(2)sin(5π/12)+log_(2)cos(5π/12)= =log_(2)(sin(5π/12)*cos(5π/12))= =log_(2)(1/2)*sin(5π/6)= =log_(2)(1/2)+log_(2)sin(5π/6) =log_(2)1/2+log_(2)(1/2)=-1-1=-2