设A=α+π/4,2A=2α+π/2
tan(2A)= tan(2α+π/2)= -cot(2α)
=-cos2α/sin2α=-b/a
tan(2A)= 2tanA/(1-tanA^2)=- b/a
解之得
[2a+根号下(4a^2+4b^2)]
tanA= -------------------------或
2b
[2a-根号下(4a^2+4b^2)]
tanA= -------------------------
2b
即:tan(α+π/4)
设A=α+π/4,2A=2α+π/2
tan(2A)= tan(2α+π/2)= -cot(2α)
=-cos2α/sin2α=-b/a
tan(2A)= 2tanA/(1-tanA^2)=- b/a
解之得
[2a+根号下(4a^2+4b^2)]
tanA= -------------------------或
2b
[2a-根号下(4a^2+4b^2)]
tanA= -------------------------
2b
即:tan(α+π/4)