题目

△abc,设向量p=(cosb/2,sinb/2),q=(cosb/2,-sinb/2)pq夹角为π/3
求角b的大小
已知tanc=根号3/2,求（sin2a·cosa-sina）/(sin2a·cos2a)

提问时间：2020-12-16

答案

令pq夹角为x ,则向量积 = p的模·q的模·cosx = 1·1·cosπ/3 = (cosb/2)·(cosb/2) + (sinb/2)·(-sinb/2) = (cosb/2)^2 - (sinb/2)^2 = cosb ,即cosb = cosπ/3 ,又∵B是三角形内角 ,∴B = π/3由tanC =√3/2 >√...