Question

在三角形ABC中，已知(sinA+sinB+sinC)(sinA+sinB-sinC)=3sinAsinB，求角C的度数

需要过程,谢谢

Answer 1

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

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

sinA+sinB+sinC=2sin[(A+B)/2]{cos[(A-B)/2]+cos[(A+B)/2]}=2sin[(A+B)/2]*2cos(A/2)cos(B/2)=4sin[(A+B)/2]cos(A/2)cos(B/2)

sinA+sinB-sinC=2sin[(A+B)/2]{cos[(A-B)/2]-cos[(A+B)/2]}=2sin[(A+B)/2]*2sin(A/2)sin(B/2)=4sin[(A+B)/2]sin(A/2)sin(B/2)

(sinA+sinB+sinC)(sinA+sinB-sinC)=16{sin[(A+B)/2]}^2*cos(A/2)cos(B/2)sin(A/2)sin(B/2)=4{sin[(A+B)/2]}^2*sinAsinB=3sinAsinB

∴{sin[(A+B)/2]}^2=3/4

cosC=cos(180-A-B)=-cos(A+B)=2*{sin[(A+B)/2]}^2-1=2*（3/4）-1=1/2

∴∠C=60°