证明：（tan(x+y)(1-tanxtany))/(tan(x-y)(1+tanxtany))=（sin(x+y))/（sin(x-y))你们是怎么想到分子分母同时乘以cosxcosy

Question

证明：（tan(x+y)(1-tanxtany))/(tan(x-y)(1+tanxtany))
=（sin(x+y))/（sin(x-y))
你们是怎么想到分子分母同时乘以cosxcosy

Answer 1

利用tan的和差化积公式有：
tan(x+y)=(tanx+tany)/(1-tanxtany);tan(x-y)=(tanx-tany)/(1+tanxtany).
所以原式
(tan(x+y)(1-tanxtany))/(tan(x-y)(1+tanxtany)) (由和差化积公式)
=[(tanx+tany)/(1-tanxtany)*(1-tanxtany)]*
[(tanx-tany)/(1+tanxtany)*(1+tanxtany)]
=(tanx+tany)/(tanx-tany) (tanx=sinx/cosx,tany=siny/cosy)
=(sinx/cosx+siny/cosy)/(sinx/cosx-siny/cosy) (分子分母同时乘以cosxcosy)
=(sinxcosy+cosxsiny)/(sinxcosy-cosxsiny) (分子分母同时用积化和差公式)
=sin(x+y)/sin(x-y)
因此所证式成立.
主要是基于两点,都属于比较基本的技巧.一个是题中把tan化成sin与cos的比,这是三角函数证明中一个切割化弦的思想,即用tan,sec等切割不好做时要考虑到化成正弦和余弦,相比之下正弦和余弦的公式也更多一点；之后是通分,虽然有些时候通分会比较麻烦,但更多时候希望得到整式的运算,而不是分式的运算,通分也会使问题变得更清楚明了.