已知 △ABC中,a,b,c为角A.B.C的对边,且a+c=2b,A-C=π/3 求sinB的值2sin[(A+C)/2] * cos[(A-C)/2] = 2sinB sin[(A+C)/2] * cos(pi/6) = sinB 这一步我看不懂,为什么1/2 cos[(A-C)/2] =cos(pi/6)
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![已知 △ABC中,a,b,c为角A.B.C的对边,且a+c=2b,A-C=π/3 求sinB的值2sin[(A+C)/2] * cos[(A-C)/2] = 2sinB sin[(A+C)/2] * cos(pi/6) = sinB 这一步我看不懂,为什么1/2 cos[(A-C)/2] =cos(pi/6)](/uploads/image/z/14337395-35-5.jpg?t=%E5%B7%B2%E7%9F%A5+%E2%96%B3ABC%E4%B8%AD%2Ca%2Cb%2Cc%E4%B8%BA%E8%A7%92A.B.C%E7%9A%84%E5%AF%B9%E8%BE%B9%2C%E4%B8%94a%2Bc%3D2b%2CA-C%3D%CF%80%2F3+%E6%B1%82sinB%E7%9A%84%E5%80%BC2sin%5B%28A%2BC%29%2F2%5D+%2A+cos%5B%28A-C%29%2F2%5D+%3D+2sinB+sin%5B%28A%2BC%29%2F2%5D+%2A+cos%28pi%2F6%29+%3D+sinB+%E8%BF%99%E4%B8%80%E6%AD%A5%E6%88%91%E7%9C%8B%E4%B8%8D%E6%87%82%EF%BC%8C%E4%B8%BA%E4%BB%80%E4%B9%881%2F2+cos%5B%28A-C%29%2F2%5D+%3Dcos%28pi%2F6%29)
已知 △ABC中,a,b,c为角A.B.C的对边,且a+c=2b,A-C=π/3 求sinB的值2sin[(A+C)/2] * cos[(A-C)/2] = 2sinB sin[(A+C)/2] * cos(pi/6) = sinB 这一步我看不懂,为什么1/2 cos[(A-C)/2] =cos(pi/6)
已知 △ABC中,a,b,c为角A.B.C的对边,且a+c=2b,A-C=π/3 求sinB的值
2sin[(A+C)/2] * cos[(A-C)/2] = 2sinB
sin[(A+C)/2] * cos(pi/6) = sinB
这一步我看不懂,为什么1/2 cos[(A-C)/2] =cos(pi/6)
已知 △ABC中,a,b,c为角A.B.C的对边,且a+c=2b,A-C=π/3 求sinB的值2sin[(A+C)/2] * cos[(A-C)/2] = 2sinB sin[(A+C)/2] * cos(pi/6) = sinB 这一步我看不懂,为什么1/2 cos[(A-C)/2] =cos(pi/6)
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(pi/6) = sinB
因为A + B + C = 180
所以:(A+C)/2 = pi/2 - B/2
所以:
cos(B/2) * √3/2 = 2sin(B/2)cos(B/2)
显然B/2不等于pi/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