已知sin(x+y)=1/3,sin(x-y)=1/5,则tanx/tany

问题描述:

已知sin(x+y)=1/3,sin(x-y)=1/5,则tanx/tany

sin(x+y)=sinxcosy+cosxsiny=1/3,sin(x-y)=sinxcosy-cosxsiny=1/5,两式作商得(sinxcosy+cosxsiny)/(sinxcosy-cosxsiny)=5/3,分子分母同时除以cosxcosy得(tanx+tany)/(tanx-tany)=5/3,所以3tanx+3tany=5tanx-5tan...