设cos^2(x-y)-cos^2(x+y)=1/2,(1+cos2x)*(1+cos2y)=1/3,则tanxtany=?
问题描述：
设cos^2(x-y)-cos^2(x+y)=1/2,(1+cos2x)*(1+cos2y)=1/3,则tanxtany=?
答
∵cos^2(x-y)-cos^2(x+y)=1/2==>[cos(x-y)+cos(x+y)][cos(x-y)-cos(x+y)]=1/2==>(2cosxcosy)(2sinxsiny)=1/2 (应用和差角公式)==>cosxcosysinxsiny=1/8.(1)(1+cos2x)*(1+cos2y)=1/3==>(2cos²x)(2cos²y)=1/...