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
网友回答
2sinB = sinA + sinC
B = 180 - A - C
2sinB = 2sin(180-A-C) = 2sin(A+C) = sinA + sinC
sinA + 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