一道高考题(08年全国卷一理科数
aCosB-bCosA=3/5c
a=2rSinA b=2rSinB c=2rSinC
2rSinACosB-2rSinBCosA=2rSinC*3/5
Sin(A-B)=3/5SinC (1)
tan(A-B)=Sin(A-B)/Cos(A-B)
=3/5SinC/√[1-Sin^2(A-B)]
=3/5SinC*5/√[(25-9Sin^2C]
=3SinC/√[25-9SinC]
SinC<=1
tan(A-B)<=3/4 tan(A-B)的最大值为3/4
。 全部
aCosB-bCosA=3/5c
a=2rSinA b=2rSinB c=2rSinC
2rSinACosB-2rSinBCosA=2rSinC*3/5
Sin(A-B)=3/5SinC (1)
tan(A-B)=Sin(A-B)/Cos(A-B)
=3/5SinC/√[1-Sin^2(A-B)]
=3/5SinC*5/√[(25-9Sin^2C]
=3SinC/√[25-9SinC]
SinC<=1
tan(A-B)<=3/4 tan(A-B)的最大值为3/4
。
收起