Question

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

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Answer 1

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