在△ABC中,设a+c=2*b,A-C=π/3,求sinB的值
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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 = 180 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
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 = 180 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
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