Question

数学三角的题

已知tan(α+β)=-2,tan(α-β)=1/2,求sin2α/sin2β的值
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Answer 1

解：指燃
tan(α+β)/tan(α-β)=sin(α+β)*cos(α-β)/[cos(α+β)sin(α-β)]=-4
sin(α+β)*cos(α-β)=-4cos(α+β)sin(α-β)
sin2α/sin2β
=sin[(α+β)+(α-β)]/握基sin[(α+β)-(α-β)]
=[sin(α+β)cos(α-β)+cos(α+β)sin(α-β)]/[sin(α+β)cos(α-β)-cos(α+β)sin(α-β)]
=[-4cos(α+β)sin(α-β)+cos(α+β)sin(α-β)]/[-4cos(α+β)sin(α-β)-cos(α+β)sin(α-β)]
=(-3)/唯皮虚(-5)
=3/5

Answer 2

tan(2α)=tan[(α+β)+(α-β)]
tan(2β)=tan[(α+β)-(α-β)]
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Answer 3

5/3