Question

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

Answer 1

答：
tan(a+b)=-2,tan(a-b)=1/2
sin2a/sin2b
=sin(a+b+a-b)/sin[a+b-(a-b)]
=[sin(a+b)cos(a-b)+cos(a+b)sin(a-b)] / [sin(a+b)cos(a-b)-cos(a+b)sin(a-b)]
上式分子分母同除以sin(a+b)cos(a-b)得下式：
=[1+tan(a-b) / tan(a+b) ] / [1-tan(a-b) / tan(a+b) ]
=[1+(1/2) / (-2) ] / [1 -(1/2) / (-2) ]
=(1-1/4)/(1+1/4)
=(3/4)/(5/4)
=3/5