Question

在三角形ABC中，已知(a+b)/a= sinB/(sinB -sinA)，且cos(A-B)+co

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

Answer 1

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}

Answer 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···