高手解题:关于解三角形1``已知A,B,C是三角形ABC的三个内角,且满足(sinA+sinB)^2-sin^2C=3s

2个回答

  • (sinA+sinB)^2-sin^2C=3sinAsinB

    sin²A+2sinAsinB+sin²B-3sinAsinB=sin²C

    sin²A-sinAsinB+sin²B=sin²(A+B)

    sin²A(sin²B+cos²B)-sinAsinB+sin²B(sin²A+cos²A)=(sinAcosB+sinBcosA)²

    展开得

    sin²A(sin²B+cos²B)-sinAsinB+sin²B(sin²A+cos²A)

    =(sinAcosB+sinBcosA)²

    =sin²Acos²B+2sinAsinBcosAcosB+sin²Bcos²A

    两边减去相同项得

    sin²Asin²B-sinAsinB+sin²Bsin²A=2sinAsinBcosAcosB

    2sin²Asin²B-sinAsinB-2sinAsinBcosAcosB=0

    sinAsinB(2sinAsinB-1-2cosAcosB)=0

    2sinAsinB-1-2cosAcosB=0

    sinAsinB-cosAcosB=1/2

    -cos(A+B)=1/2

    cos(A+B)=-1/2

    A+B=120°

    2.∵acosA+bcosB=ccosC

    ∴sinAcosA+sinBcosB=sinCcosC

    ∴sin2A+sin2B=sin2C=sin(2π-2A-2B)=-sin(2A+2B)

    ∴0=sin2A+sin2B+sin(2A+2B)

    =sin2A+sin2B+sin2Acos2B+sin2Bcos2A

    =sin2A(1+cos2B)+sin2B(1+cos2A)

    =4sinAcosA(cosB)^2+4sinBcosB(cosA)^2

    =4cosAcosBsin(A+B)

    ∵sin(A+B)=sin(π-C)=sinC>0

    ∴cosA=0或cosB=0

    ∴A=π/2或B=π/2

    ∴△ABC是以a或b为斜边的直角三角形