在三角形ABC中,已知a+c=2b,A-C=60度,求sin(B/2) (a,b,c为边,A,B,C为角)

题目

在三角形ABC中,已知a+c=2b,A-C=60度,求sin(B/2) (a,b,c为边,A,B,C为角)

答案

在三角形ABC中,
a/sinA = b/sinB = c/sinC
a + c = 2b
sinA + sinC = 2sinB
2sin[(A+C)/2] * cos[(A-C)/2] = 2 * 2 * sin(B/2) * cos(B/2)
sin[(A+C)/2] * cos30 = 2 * sin(B/2) * cos(B/2)
cos[90 - (A+C)/2] * cos30 = 2 * sin(B/2) * cos(B/2)
cos(B/2) * cos30 = 2 * sin(B/2) * cos(B/2)
cos30 = 2 * sin(B/2)
sin(B/2) = 根3 / 4