已知(sinα-sinβ)/[sin(α-β)]=a,(cosα-cosβ)/[sin(α+β)]=b,求sin(α-β
1个回答

(sinα-sinβ)/[sin(α-β)]=a,.(1)

(cosα-cosβ)/[sin(α+β)]=b,.(2)

(1)*(2)得

[sinαcosα-sinαcosβ-cosαsinβ+sinβcosβ]/sin(α-β)sin(α+β)=ab

[1/2(sin2α+sin2β)-sin(α+β)]/sin(α-β)sin(α+β)=ab

[sin(α+β)cos(α-β)-sin(α+β)]/sin(α-β)sin(α+β)=ab

sin(α+β)[cos(α-β)-1]/sin(α-β)sin(α+β)=ab

[cos(α-β)-1]/sin(α-β)=ab

cos(α-β)-1=absin(α-β)

cos^2(α-β)=[1+absin(α-β)]^2

1-sin^2(α-β)=1+2absin(α-β)+(ab)^2sin^2(α-β)

sin(α-β){[(1+(ab)^2]sin(α-β)+2ab}=0

sin(α-β)不等于0

[1+(ab)^2]sin(α-β)+2ab=0

sin(α-β)= -2ab/[1+(ab)^2]