Question

如何化简函数4[sin^2（B+C）/2]-cos2A=7/2

Answer 1

∵(B+C)/2=π/2-A/2

sin[（B+C）/2]=cos(A/2)

sin^2[（B+C）/2]=cos²A/2=(1+cosA)/2

cos2A=2cos²A-1

∴4[sin^2（B+C）/2]-cos2A=2(1+cosA)-2cos²A+1=7/2

4cos²A-4cosA+1=0

(2cosA-1)²=0,cosA=1/2

A=π/3