已知：2tan2β＝tanα + tanβ求证:tan(α-β)=sin2β

硬算
2tan2β＝tanα + tanβ

所以 tanα=2tan2β-tanβ = 4tanβ/(1-tan^2 β)-tanβ
=(3tanβ+tan^3 β)/(1-tan^2 β)

左边= tan(α-β)

= (tanα - tanβ)/(1+tanα * tanβ)

= [(3tanβ+tan^3 β)/(1-tan^2 β) - tanβ] /
[1+(3tanβ+tan^3 β)/(1-tan^2 β) * tanβ]

= [(2tanβ+2tan^3 β)/(1-tan^2 β)] /
[(1-tan^2 β + 3tan^2 β + tan^4 β )/(1-tan^2 β)]

= (2tanβ+2tan^3 β)/(1+2tan^2 β + tan^4 β )

= [2tanβ*(1+2tan^2 β)]/[(1+tan^2 β)^2]

= 2tanβ/(1+tan^2 β)

= sin2β=右边
原命题的证