在△ABC中,已知8sin^2[(B+C)/2]-2cos2A=7

1、
8[1-cos(B+C)]/2-2cos2A=7
4[1-cos(180-A]-2(2cos²A-1)=7
4+4cosA-4cos²A+2=7
4cos²A-4cosA+1=0
cosA=1/2
A=π/3
2、
B+C=2π/3
C=2/3-B
sinB+sinC
=sinB+sin(2π/3-B)
=sinB+sin2π/3cosB-cos2π/3sinB
=3/2*sinB+√3/2*cosB
=√3(sinB√3/2+cosB*1/2)
=√3(sinBcosπ/6+cosBsinπ/6)
=√3sin(B+π/6)
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