在三角形ABC中,角A,B,C的对边分别为a,b,

在三角形ABC中,角A,B,C的对边分别为a,b,c,已知向量m=(2a-c,b)与向量n=(cosB,-cosC)互相

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1)m∥n,∴b/(2a-c)=cosB/cosC正弦定理:a/sinA=b/sinB=c/sinC∴b/(2a-c)=sinB/(2sinA-sinC)=cosB/cosC∴2sinAcosB=sinCcosB+sinBcosC=sin(B+C)=sin(180°-A)=sinA∴cosB=1/2,B=60°2)C=120°-Ay=cosA+cosC=(1+cos2A)/2+(1+cos2C)/2=1+(1/2)[cos2A+cos2(120°-A)] =1+(1/2)[cos2A+cos(240°-2A)]=1+(1/2)[cos2A-(1/2)cos2A-(√3/2)sin2A] =1+(1/2)[(1/2)cos2A-(√3/2)sin2A]=1+(1/2)cos(2A+60°)∵0°