已知锐角三角形ABC中,sin(A+B)=3/5,sin(A-B)=1/5.1求证:tanA=2tanB.2设AB=3.

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  • sin(A+B)=3/5,

    sin(A-B)=1/5

    则:sin(A+B)=3sin(A-B)

    sinAcosB+cosAsinB=3sinAcosB-3cosAsinB

    2sinAcosB=4cosAsinB

    sinA/cosA=2sinB/cosB

    tanA=2tanB

    (2)sin(A+B)=3/5

    sinC=sin[180°-(A+B)]=sin(A+B)=3/5

    则:cosC=根号(1-sin²C)=4/5

    则:tanC=sinC/cosC=3/4

    tan(A+B)=tan(180°-C)=-tanC=-3/4

    (tanA+tanB)/(1-tanAtanB)=-3/4

    3tanB/(1-2tan²B)=-3/4

    2tan²B-1=4tanB

    2tan²B-4tanB-1=0

    tanB=(4+2√6)/4=(2+√6)/2

    tanA=2tanB=2+√6

    设AB边上的高是CD=H

    tanA=H/AD,tanB=H/BD

    AD+BD=AB=H/tanA+H/tanB=H/tanA+2H/tanA=3H/tanA=3

    H=tanA=2+根号6

    即AB边上的高是;2+根号6.