已知b－c=2a*cos(π/3 +C),结合正弦定理 a/sinA=b/sinB=c/sinC 得：sinB-sinC=2sinA·cos(π/3 +C),→sin(A+C)-sinC=2sinA·[cos(π/3)cosC-sin(π/3)sinC],→sinAcosC+cosAsinC-sinC=2sinA(1/2·cosC-√3/2·sinC),...