问题
解答题
已知tan(α-β)=
|
答案
证明:∵tan(α-β)=
,tanα-tanβ 1+tanαtanβ
=sin2β 5-cos2β
=2sinβcosβ 5-(1-2sin2β)
=2sinβcosβ 4+2sin2β sinβcosβ 2+sin2β
=
=sinβcosβ 2cos2β+3sin2β
=tanβ 2+3tan2β
,
tanβ-tanβ3 2 1+
tan2β3 2
且tan(α-β)=
,sin2β 5-cos2β
∴
=tanα-tanβ 1+tanαtanβ
,
tanβ-tanβ3 2 1+
tan2β3 2
则tanα=
tanβ,即2tanα=3tanβ.3 2