因为向量a,b垂直则a*b=0,cosθ+2sinθ-2cosθ=0得tanθ=1/2
tan2θ=2tanθ/(1-tan^2θ)=1/(1-1/4)=4/3
tan(2θ+π/4)=(tan2θ+1)/(1-tan2θ)=(4/3+1)/(1-4/3)=-7
因为向量a,b垂直则a*b=0,cosθ+2sinθ-2cosθ=0得tanθ=1/2
tan2θ=2tanθ/(1-tan^2θ)=1/(1-1/4)=4/3
tan(2θ+π/4)=(tan2θ+1)/(1-tan2θ)=(4/3+1)/(1-4/3)=-7