Question

已知tanα=4根号3，cos（α+β）=-11/14，α，β均为锐角，求cosβ的值

Answer 1

{A = arctan(4*sqrt(3)), B = 1/3*Pi}, {A = arctan(4*sqrt(3)), B = -arctan(39/71*sqrt(3))+Pi}, {A = arctan(4*sqrt(3))-Pi, B = -2/3*Pi}, {A = arctan(4*sqrt(3))-Pi, B = -arctan(39/71*sqrt(3))};所以cosB=0.5。

Answer 2

∵ tanβ=sinβ/cosβ
∴(sinβ)^2=48(1-sin^2β) sinβ=4√3/7
∴cosβ=√(1-sin^2β)=1/7
∵sin(α+β)=√[1-cos^2(α+β)]=5√3/14
∴cosα=cos(α+β-β)=cos(α+β)cosβ+sin(α+β)sinβ=1/2
tanα=4√3
sinα=4√3/7
cosα=1/7

cos(α+β)=-11/14，且α、β均为锐角
0<α+β<π
sin(α+β)=5√3/14

cosβ
=cos[(α+β)-α]
=cos(α+β)cosα+sin(α+β)sinα
=-11/14*1/7+5√3/14*4√3/7
=-11/98+60/98
=49/98
=1/2