在三角形ABC中,已知(a+b)/a= sinB/(sinB -sinA),且cos(A-B)+cosC=1-cos2C(1)试确定三角形的形状(2)求(a+ c)/b的取值范围
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![在三角形ABC中,已知(a+b)/a= sinB/(sinB -sinA),且cos(A-B)+cosC=1-cos2C(1)试确定三角形的形状(2)求(a+ c)/b的取值范围](/uploads/image/z/3689896-40-6.jpg?t=%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2C%E5%B7%B2%E7%9F%A5%28a%2Bb%29%2Fa%3D+sinB%2F%28sinB+-sinA%29%2C%E4%B8%94cos%28A-B%29%2BcosC%3D1-cos2C%281%29%E8%AF%95%E7%A1%AE%E5%AE%9A%E4%B8%89%E8%A7%92%E5%BD%A2%E7%9A%84%E5%BD%A2%E7%8A%B6%282%29%E6%B1%82%28a%2B+c%29%2Fb%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
在三角形ABC中,已知(a+b)/a= sinB/(sinB -sinA),且cos(A-B)+cosC=1-cos2C(1)试确定三角形的形状(2)求(a+ c)/b的取值范围
在三角形ABC中,已知(a+b)/a= sinB/(sinB -sinA),且cos(A-B)+cosC=1-cos2C
(1)试确定三角形的形状
(2)求(a+ c)/b的取值范围
在三角形ABC中,已知(a+b)/a= sinB/(sinB -sinA),且cos(A-B)+cosC=1-cos2C(1)试确定三角形的形状(2)求(a+ c)/b的取值范围
根据正弦定理,(a+b)/a= sinB/(sinB -sinA) =(sinA+sinB)/sinA
∴sinA·sinB = (sinB+sinA)(sinB-sinA)
= 2sin[(B+A)/2]·cos[(B-A)/2]·2·cos[(B+A)/2]·sin[(B-A)]
=sin(B-A)·sin(B+A)
=sinC·sin(B-A)
cos(A-B)+cosC=1-cos2C 即 2sinA·sinB = 2(sinC)^2,∴(sinC)^2 = sinA·sinB
∴(sinC)^2 = sinC·sin(B-A),∴cosB·sinA = 0,∵sinA≠0,∴cosB = 0,∴B =π/2
∴△ABC是以B为直角的Rt△
又∵sinB/(sinB -sinA) =(sinA+sinB)/sinA,∴1/[1 - sinA] = (1 + sinA)/sinA
∴sinA = 1-(sinA)^2,解得sinA = (√5 - 1)/2 ,∴sinC = cosA = [(2√5 - 2)^(1/2)]/2
∴(a+ c)/b = (sinA + sinC)/sinB = sinA + sinC = (√5 - 1)/2 + {[(2√5 - 2)^(1/2)]/2}
先用正弦定理将a,b,c化成sinA,sinnB,sinC的形式。然后化简···
第二问,通过二倍角公式,cos2c=(cosc)平方-1,得cos(A-B) ···不会打!找不到符号···