设tan(α+β)=2,tan(α-β)=5,求tan2α,tan2β,tan4α,tan4β

设tan(α+β)=2,tan(α-β)=5,求tan2α,tan2β,tan4α,tan4β
设tan(α+β)=2,tan(α-β)=5,求tan2α,tan2β,tan4α,tan4β

设tan(α+β)=2,tan(α-β)=5,求tan2α,tan2β,tan4α,tan4β
tan2α
=(2+5)/(1-2*5)
=-10/9
tan2β
=(2-5)/(1+2*5)
=-3/11
tan4α
=2*(-10/9)/(1-(-10/9)²)
=180/19
tan4β
=2*(-3/11)/(1-(-3/11)²)
=-33/56

tan2a=tan(α+β+a-β)=[tan(α+β)+tan(α-β)]/[1-tan(α-β)tan(α+β)]=-7/9
tan2β=tan(α+β-a+β)=[tan(α+β)-tan(α-β)]/[1+tan(α-β)tan(α+β)]=-3/11
tan4α=-63/65
tan4β=-33/65