Question

已知a-β=π/6，tanα-tanβ=3，求cos(α+β)

Answer 1

α-β=π/6，tanα-tanβ=3，求cos(α+β)
solution
tanα-tanβ=3

(sinα.cosβ - cosα.sinβ) / (cosα.cosβ)=3
sin(α-β) =3cosα.cosβ
1/2 =3cosα.cosβ
cosα.cosβ = 3/2 (1)
tan(α-β)=tan(π/6)
(tanα-tanβ)/(1+tanα.tanβ) = √3/3
3/(1+tanα.tanβ) = √3/3
1+tanα.tanβ = 3√3
tanα.tanβ = 3√3 -1
sinα.sinβ/(cosα.cosβ) = 3√3 -1
sinα.sinβ =(3/2)(3√3 -1) (2)
cos(α+β)

=cosα.cosβ - sinα.sinβ
=3/2 - (3/2)( 3√3 -1)
=3 - (9/2)√3

Answer 2