△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)
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![△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)](/uploads/image/z/3976715-11-5.jpg?t=%E2%96%B3abc%2C%E8%AE%BE%E5%90%91%E9%87%8Fp%3D%28cosb%2F2%2Csinb%2F2%29%2Cq%3D%28cosb%2F2%2C-sinb%2F2%29pq%E5%A4%B9%E8%A7%92%E4%B8%BA%CF%80%2F3%E6%B1%82%E8%A7%92b%E7%9A%84%E5%A4%A7%E5%B0%8F%E5%B7%B2%E7%9F%A5tanc%3D%E6%A0%B9%E5%8F%B73%2F2%2C%E6%B1%82%EF%BC%88sin2a%C2%B7cosa-sina%EF%BC%89%2F%28sin2a%C2%B7cos2a%29)
△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 .