Question

1+cos2A-cos2B-cos2C=2sinBsinC+1-

Answer 1

1+cos2A-cos2B-cos2C=2sinBsinC
1+1-2sin²A-1+2sin²B-1+2sin²C=2sinBsinC
sin²B+sin²C-sin²A=sinBsinC
sin²B+sin²C-sin²[180°-(B+C)]=sinBsinC
sin²B+sin²C-sin²(B+C)=sinBsinC
sin²B+sin²C-(sinBcosc+cosBsinC)²=sinBsinC
sin²B+sin²C-(sin²Bcos²c+cos²Bsin²C+2sinBsinCcosBcosC)=sinBsinC
sin²B(1-cos²C)+sin²C(1-cos²B)-sinBsinC(1+2cosBcosC)=0
2sin²Bsin²C-sinBsinC(1+2cosBcosC)=0
sinBsinC(2sinBsinC-2cosBcosC-1)=0
sinBsinC[-2cos(B+C)-1]=0
sinBsinC[2cos(180°-B-C)-1]=0
sinBsinC(2cosA-1)=0
∵A、B、C均属于（0°，180°）
∴sinBsinC≠0
∴(2cosA-1)=0
cosA=0.5
∴ A=60°