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问题：已知tan(α+β)=-2,tan(α-β)=1/2,求sin2α/sin2β的值
答案：↓↓↓ 戴余良的回答： 网友采纳 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 | 
