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
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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
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
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
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/...
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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 = 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
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