2.在锐角△ABC中,若cos²A,cos²B,cos²C的和等于sin²A,s

3个回答

  • 不失一般性地设:(cosA)^2+(cosB)^2+(cosC)^2=(sinC)^2,则:

    (cosA)^2+(cosB)^2+2(cosC)^2=1,

    ∴2(cosA)^2+2(cosB)^2+4(cosC)^2=2,

    ∴[2(cosA)^2-1]+[2(cosB)^2-1]+2(cosC)^2=-2(cosC)^2,

    ∴cos2A+cos2B+2(cosC)^2=-2(cosC)^2,

    ∴2cos(A+B)cos(A-B)+2[cos(A+B)]^2=-2(cosC)^2,

    ∴2cos(A+B)[cos(A+B)+cos(A-B)]=-2(cosC)^2,

    ∴-2cosC(2cosAcosB)=-2(cosC)^2,

    ∴2cosAcosB=cosC,

    ∴2sinCcosAcosB=sinCcosC,

    ∴2(sinC/cosC)cosAcosB=sinC,

    ∴2tanCcosAcosB=sin(A+B),

    ∴2tanCcosAcosB=sinAcosB+cosAsinB,

    ∴2tanC=(sinA/cosA)+(sinB/cosB),

    ∴2tanC=tanA+tanB,

    ∴tanA、tanC、tanB组成一个等差数列.

    同理可证:

    (cosA)^2+(cosB)^2+(cosC)^2=(sinA)^2 时,tanB、tanA、tanC成等差数列.

    (cosA)^2+(cosB)^2+(cosC)^2=(sinB)^2 时,tanA、tanB、tanC成等差数列.