已知tan(π/4+θ)=2,求sin2θ+cos2θ+1的值.

tan(π/4+θ)
=(tanπ/4+tanθ)/(1-tanπ/4tanθ)
=(1+tanθ)/(1-tanθ)
=2
由此可以解出 tan θ= 1/3.
万能公式
cos2θ＝(1-tan^θ)/(1+tan^θ)
sin2θ＝ 2tanθ/(1+tan^θ)
cos2θ+sin2θ
=(1-tan^θ)/(1+tan^θ) +2tanθ/(1+tan^θ)
=(1-tan^θ+2tanθ)/(1+tan^θ)
=[1-1/9+2*1/3]/(1+1/9)
=7/5
sin2θ+cos2θ+1=7/5+1=12/5
^表平方

tan(π/4+θ)=(tanπ/4+tanθ)/(1-tanπ/4*tanθ)=(1+tanθ)/(1-tanθ)=2 解得tanθ=1/3
所以 sinθ=1/√10，cosθ=3/√10
sin2θ+cos2θ+1=2sinθcosθ+2cosθ*cosθ-1+1=2cosθ(sinθ+cosθ)
=2*3/√10*(1/√10+3/√10)=12/5

tan(π/4+θ)=[tan(π/4) + tanθ] / [1 - tan(π/4)tanθ] =(1 + tanθ) / (1 - tanθ) =2
tanθ=1/3
sin2θ + cos2θ + 1
=(2tanθ)/[1+(tanθ)^2] + [1-(tanθ)^2]/[1+(tanθ)^2] + 1
=12/5