tanα=4√3,cos(α+β)=-11/14,求sinβ和sin(α+2β)

题目

tanα=4√3,cos(α+β)=-11/14,求sinβ和sin(α+2β)

答案

有第一条件,求出sinα=4√3/7 cosα=1/7
cos(α+β)=cosαcosβ- sinαsinβ = -11
又 sinβ^2+cosβ^2 = 1;
求出 cosβ = 1/2
sinβ = -√3/2
β = 240°
2β = 480°
sin(α+2β) = sinαcos2β+cosαsin2β = -11/14