Question

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

Answer 1

∵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