已知三角形ABC中,sinA,sinB ,sinC是等差数列,求证2cos（A+C）/2=cos（A-C）/2

2sinB = sinA + sinCB = 180 - A - C2sinB = 2sin(180-A-C) = 2sin(A+C) = sinA + sinCsinA + sinC = 2 * sin[(A+C)/2] * cos[(A-C)/2]2sin(A+C) = 4 * sin[(A+C)/2] * cos[(A+C)/2]so,2cos（A+C）/2=cos（A-C）/2...