Question

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

Answer 1

(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]