一道三角函数题目
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少条件吧?"a+c=2b"
正弦定理得:
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
正弦定理得:
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
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