△ABC中,cos(A-C)+cosB=3/2,b方=ac,求B.问:要不要舍去一个值啊?

问题描述:

△ABC中,cos(A-C)+cosB=3/2,b方=ac,求B.问:要不要舍去一个值啊?

cos (A-C) + cos B = 3/2所以 cos(A-C) + cos(pai - A - C) = 3/2cos(A-C) + cos(pai - A - C) = cos(A - C ) - cos(A + C) = cosAcosC + sinAsinC - cosAcosC + sinAsinC = 2sinAsinC = 3/2所以 sinAsinC = 3/4正弦...