锐角三角形ABC满足tanA-1/sin2A=tanB ,那么cos(A-B/2)=___________若C=30°,则A=__...锐角三角形ABC满足tanA-1/sin2A=tanB ,那么cos(A-B/2)=___________若C=30°,则A=_______
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![锐角三角形ABC满足tanA-1/sin2A=tanB ,那么cos(A-B/2)=___________若C=30°,则A=__...锐角三角形ABC满足tanA-1/sin2A=tanB ,那么cos(A-B/2)=___________若C=30°,则A=_______](/uploads/image/z/5215896-0-6.jpg?t=%E9%94%90%E8%A7%92%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E6%BB%A1%E8%B6%B3tanA-1%2Fsin2A%3DtanB+%2C%E9%82%A3%E4%B9%88cos%EF%BC%88A-B%2F2%EF%BC%89%3D___________%E8%8B%A5C%3D30%C2%B0%2C%E5%88%99A%3D__...%E9%94%90%E8%A7%92%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E6%BB%A1%E8%B6%B3tanA-1%2Fsin2A%3DtanB+%2C%E9%82%A3%E4%B9%88cos%EF%BC%88A-B%2F2%EF%BC%89%3D___________%E8%8B%A5C%3D30%C2%B0%2C%E5%88%99A%3D_______)
锐角三角形ABC满足tanA-1/sin2A=tanB ,那么cos(A-B/2)=___________若C=30°,则A=__...锐角三角形ABC满足tanA-1/sin2A=tanB ,那么cos(A-B/2)=___________若C=30°,则A=_______
锐角三角形ABC满足tanA-1/sin2A=tanB ,那么cos(A-B/2)=___________若C=30°,则A=__...
锐角三角形ABC满足tanA-1/sin2A=tanB ,那么cos(A-B/2)=___________若C=30°,则A=_______
锐角三角形ABC满足tanA-1/sin2A=tanB ,那么cos(A-B/2)=___________若C=30°,则A=__...锐角三角形ABC满足tanA-1/sin2A=tanB ,那么cos(A-B/2)=___________若C=30°,则A=_______
因tanA-(1/sin2A)=tanB
化简、整理可得-cos2A/sin2A=sinB/cosB
即-cos2A*cosB=sin2A*sinB
则cos2A*cosB+sin2A*sinB=0
也就是cos(2A-B)=0
(1)根据半角公式cos(A-B/2)= √〔/2〕
所以cos(A-B/2)=√2/2
(2)依cos(A-B/2)=√2/2可得,A-B/2=45度
因已知C=30度,且三角形内角和A+B+C=180度
所以A=80度
tanA-(1/sin2A)=tanB
-cos2A/sin2A=sinB/cosB
-cos2A*cosB=sin2A*sinB
cos2A*cosB+sin2A*sinB=0
cos(2A-B)=0
根据半角公式cos(A-B/2)= √〔<1+cos(2A-B)>/2〕
cos(A-B/2)=√2/2
cos(A-B/2)=√2/2
A-B/2=45度
因已知C=30度,且三角形内角和A+B+C=180度
所以A=80度
二分之根号二,80度