请教一道高一数学题“已知tan（θ－π/4）=2,求4cos²θ－3sin2θ的值”

tan(a-π/4)=(tana-1)/(1+tana)=2
tana=-3
4cos^2a-3sin2a=4cos^2a-6sinacosa
=4cos^2a-6sinacos^2a/cosa
=2cos^2a(2-3tana)=22cos^2a=22/(1+tan^2a)=22/10=11/5

tan(θ-π/4)=2.===>(tanθ-1)/[1+tanθ]=2===>tanθ=-3.4(cosθ)^2-3sin2θ=[4(cosθ)^2-6sinθcosθ]/[(sinθ)^2+(cosθ)^2]=[4-6tanθ]/[1+(tanθ)^2](此处上下同除以（tanθ)^2)=22/10=11/5.