在△ABC中,求证:sin^2A+sin^2B+cos^2C+2sinAsinBcos(A+B)=1

3个回答

  • 由题意:1-sin^2A=cos^2A

    sin^2B+cos^2C+2sinAsinBcos(A+B)=

    =sin^2B+cos^2C-2sinAsinBcosC

    =sin^2B +cosC(cosC-2sinAsinB)

    =sin^2B -cosC[cos(A+B)+2sinAsinB]

    =sin^2B-cosC[cosAcosB-sinAsinB+2sinAsinB]

    =sin^2B-cosC[cosAcosB+sinAsinB]

    =sin^2B +cos(A+B)(cosAcosB+sinAsinB)

    =sin^2B+(cosAcosB-sinAsinB)(cosAcosB+sinAsinB)

    =sin^2B +cos^2Acos^2B -sin^2Asin^2B

    =sin^2B(1-sin^2A)

    =sin^2Bcos^2A +cos^2Bcos^2A

    =cos^2A

    所以sin^2A+sin^2B+cos^2C+2sinAsinBcos(A+B)=1