Question

在三角形ABC中，求证sinA+sinB+sinC=4cosA/2cosB/2cosC/2.

Answer 1

4cos(A/2)cos(B/2)cos(C/2)=4cos(A/2)cos(B/2)cos(pi/2-A/2-B/2)=4cos(A/2)cos(B/2)sin(A/2+B/2)
=4cos(A/2)cos(B/2)(sin(A/2)cos(B/2)+sin(B/2)cos(A/2))
=2sinAcos(B/2)^2+2sinBcos(A/2)^2
==>
4cos(A/2)cos(B/2)cos(C/2)-sinA-sinB=sinA(2cos(B/2)^2-1)+sinB(2cos(A/2)^2-1)
=sinAcosB+sinBcosA=sin(A+B)=sin(pi-A-B)=sin(A+B)
==>
sinA+sinB+sinC=4cos(A/2)cos(B/2)cos(C/2)