问一道数学题目，关于三角函数的

tan(π/4+θ)
=(tanπ/4+tanθ)/(1-tanπ/4*tanθ)
=(1+tanθ)/(1-tanθ)
=(1+2)/(1-2)
=-3

cos2θ
=cos²θ-sin²θ
=(cos²θ-sin²θ)/(cos²θ+sin²θ)
=(1-tan²θ)/(1+tan²θ)
=(1-2²)/(1+2²)
=-3/5

tan（π/4+θ）
=(tanπ/4+tanθ）)/(1-tanπ/4tanθ）
=(1+2)/(1-1*2)
=-3

由万能公式
cos2θ
=[1-(tanθ)^2/(1+(tanθ)^2]
=(1-4)/(1+4)
=-3/5

tan（π/4+θ）=(tanθ+1)/(tanθ-1)=3
tanθ=2 ,cosθ^2=1/5
cos2θ=2cosθ^2-1=-3/5

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

用公式把它展开成分数的样子，就好了。第二个是还是用分数的公式求出tan2o，然后sin20＝tan×sin