在三角形ABC中,已知ln（sinA+sinB）=lnsinA+lnsinB-ln（sinB-sin

在三角形ABC中,已知ln（sinA+sinB）=lnsinA+lnsinB-ln（sinB-sin

1,1-cos2C = cos(A+B) + cosC =0cos2C = 1C =π/2直角三角形2,ln(sinA+sinB)=lnsinA+ln(sinB-sinA)+lnsinBsinA+sinB = sinAsinB(sinB-sinA) (a+b)/b = (sinA+sinB)/sinB = sinA(sinB-sinA)注意sinB-sinA>0所以(a+b)/b>0(a+b)/b = sinA(cosA-sinA)=1/2sin2A - 1/2(1-cos2A)=1/2(sin2A+cos2A) -1/2≤√2/2-1/2当A=π/8时取得所以0======以下答案可供参考======供参考答案1:不对啊供参考答案2:（1）ln(sinA+sinB)=ln(sinA)+ln(sinB)-ln(sinB-sinA)所以可得ln(sinA+sinB)=ln[sinAsinB/(sinB-sinA)]所以sinA+sinB=sinAsinB/(sinB-sinA)即(sinB)^2-(sinA)^2=sinAsinB，根据正弦定理a/sinA=b/sinB=c/sinC可得sinA:sinB=a:b所以可得b^2-a^2=abcos(A-B)+cosC=1-cos2C=2(sinC)^2因为A+B+C=π，所以C=π-(A+B)所以cosC=-cos(A+B)所以cos(A-B)-cos(A+B)=2(sinC)^2化简可得2sinAsinB=2(sinC)^2再根据正弦定理可得：ab=c^2所以可得b^2-a^2=c^2即a^2+c^2=b^2所以是直角三角形（2）根据正弦定理可得：(a+c)/b=(sinA+sinC)/sinB=sinA+sinC=sinA+cosA=√2sin(A+π/4)0所以1

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