求详细过程，非常感谢！

解题思路： 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/
解题过程：

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/