已知tan(α+β)=2tan(α-β),求(sin2β)/(sin2α)的值.

1个回答

  • 希望LZ有耐心看下去:

    tan(α+β)=2tan(α-β)

    tan(α+β)/tan(α-β)=2

    [sin(α+β)/cos(α+β)]/[sin(α-β)/cos(α-β)]=2

    [sin(α+β)/cos(α+β)]*[cos(α-β)/sin(α-β)]=2

    [(sinαcosβ+cosαsinβ)*(cosαcosβ+sinαsinβ)]/

    [(cosαcosβ-sinαsinβ)*(sinαcosβ-cosαsinβ)]=2

    [sinαcosαcos^2(β)+sin^2(α)sinβcosβ+cos^2(β)sinβcosβ+

    sinαcosαsin^2(β)]/[sinαcosαcos^2(β)-cos^2(α)sinβcosβ-

    sin^2(α)sinβcosβ+sinαcosαsin^2(β)]=2

    {sinαcosα[cos^2(β)+sin^2(β)]+sinβcosβ[sin^2(α)+cos^2(α)]}

    /{sinαcosα[cos^2(β)+sin^2(β)]-sinβcosβ[cos^2(α)+sin^2(α)]}=2

    (sinαcosα+sinβcosβ)/(sinαcosα-sinβcosβ)=2

    (sin2α/2+sin2β/2)/(sin2α/2-sin2β/2)=2

    (sin2α+sin2β)/(sin2α-sin2β)=2

    sin2α+sin2β=2sin2α-2sin2β

    3sin2β=sin2α

    sin2β/sin2α=1/3