在△ABC中,求证sinA+sinB-sinC=4sinA/2*sinB/2*cosC/2

1个回答

  • sinA+sinB

    =sin[(A+B)/2+(A-B)/2]+sin[(A+B)/2-(A-B)/2]

    =sin(A+B)/2cos(A-B)/2+cos(A+B)/2sin(A-B)/2+sin(A+B)/2cos(A-B)/2-cos(A+B)/2sin(A-B)/2

    =2sin(A+B)/2cos(A-B)/2

    A+B=180-C

    所以左边=2sin(90-C/2)cos(A-B)/2-2sinC/2cosC/2

    =2cosC/2cos(A-B)/2-2sinC/2cosC/2

    =2cosC/2[cos(A-B)/2-sinC/2]

    =2cosC/2[cos(A-B)/2-sin(180-A-B)/2]

    =2cosC/2[cos(A-B)/2-sin(90-A/2-B/2)]

    =2cosC/2[cos(A/2-B/2)-cos(A/2+B/2)]

    =2cosC/2(cosA/2cosB/2+sinA/2sinB/2-cosA/2cosB/2+sinA/2sinB/2)

    =4cosCsinA/2sinB/2=右边

    命题得证