Question

已知△ABC中，a，b，c分别为A,B,C的对边，且a+c=2b，A-C=π/3，求sinB的值

已知△ABC中，a，b，c分别为A,B,C的对边，且a+c=2b，A-C=π/3，求sinB的值
请尽量解释清楚！拜托！谢谢！

Answer 1

sinA +sinC = 2sinB
2sin[(A+C)/2] * cos[(A-C)/2] = 2sinB
sin[(A+C)/2] * cos(π/6) = sinB
因为A + B + C = π
所以：(A+C)/2 = π/2 - B/2
cos(B/2) * √3/2 = 2sin(B/2)cos(B/2)
显然B/2不等于π/2，cos(B/2)不等于0
所以：
sin(B/2) = √3/4
cos(B/2) = √13/4
sinB = 2sin(B/2)cos(B/2) = √39/8

Answer 2

因为 a + c = 2b
由正弦定理，知：
sinA +sinC = 2sinB
2sin[(A+C)/2] * cos[(A-C)/2] = 2sinB
sin[(A+C)/2] * cos(π/6) = sinB
因为A + B + C = π
所以：(A+C)/2 = π/2 - B/2
所以：
cos(B/2) * √3/2 = 2sin(B/2)cos(B/2)
显然B/2不等于π/2，cos(B/2)不等于0
所以：
sin(B/2) = √3/4
cos(B/2) = √13/4
sinB = 2sin(B/2)cos(B/2) = √39/8