△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)
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/28 10:50:59
△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)
△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)
△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)
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 >√3/3 ,故C > π/6 ,故A < π/2 ,∴△ABC是锐角三角形 ,即sin A≠0 ,∴(sin2A·cosA-sinA)/(sin2A·cos2A)= [sinA·(2cosA·cosA - 1)]/[sinA·2cosA·cos2A] = [2(cosA)^2 - 1]/[2cosA·cos2A] = cos2A/[2cosA·cos2A] = 1/(2cosA) ,由tanC = √3/2 ,代入同角关系式可得:sinC = (√21)/7 ,cosC = (2√7)/7 ,又cosB = 1/2 ,sinB = √3/2 ,∴cosA = cos[π -(B+C)]= -cos(B+C)= sinB·sinC - cosB·cosC =1/(2√7),
∴1/(2cosA) = √7 ,即 (sin2A·cosA-sinA)/(sin2A·cos2A) = √7 .