若1+tanx|1-tanx=2008,则1|cos2x+tan2x=?
1个回答

1/cos2x+tan2x=(1+sin2x)/cos2x

=(sin²x+cos²x+2sinxcosx)/(cos²x-sin²x)

=(sinx+cosx)²/[(cosx+sinx)(cosx-sinx)]

=(sinx+cosx)/(cosx-sinx)

=(tanx+1)/(1-tanx)

=2008

sinx+siny=sin225°=sin(180°+45°)=-sin45°=-√2/2

cosx+cosy=cos225°=cos(180°+45°)=-cos45°=-√2/2,

那么(sinx+siny)²=sin²x+2sinxsiny+sin²y=1/2

(cosx+cosy)²=cos²x+2cosxcosy+cos²y=1/2

两式相加,得:1+1+2(cosxcosy+sinxsiny)=1

那么2+2cos(x-y)=1

所以cos(x-y)=-1/2