设cos^2(x-y)-cos^2(x+y)=1/2,(1+cos2x)*(1+cos2y)=1/3,则tanxtany=?

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设cos^2(x-y)-cos^2(x+y)=1/2,(1+cos2x)*(1+cos2y)=1/3,则tanxtany=?
设cos^2(x-y)-cos^2(x+y)=1/2,(1+cos2x)*(1+cos2y)=1/3,则tanxtany=?

设cos^2(x-y)-cos^2(x+y)=1/2,(1+cos2x)*(1+cos2y)=1/3,则tanxtany=?
∵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)=1/3 (应用倍角公式)
==>cos²xcos²y=1/12.(2)
∴由(1)式和()式,得tanxtany=(sinx/cosx)(siny/cosy)
=(sinxsiny)/(cosxcosy)
=(cosxcosysinxsiny)/(cos²xcos²y) (分子分母同乘cosxcosy)
=(1/8)/(1/12)
=3/2

=(sin2x+cos2x)/2+3/2 =(√2/2)*sin(2x+π/4)+3/2 所以T=2π2、（1）(a+c)^2=15cos^2 x+1+15sin^2 y+1+8(cosxcosy-sinx

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