cosa=1/7
sinα = 4√3/7,
cos(a+β)=-11/14
sin(α+β)= 5√3/14,
cos(α+β)=cosαcosβ - sinαsinβ
= 1/7*cosβ - 4√3/7*sinβ
= -11/14,

sin(α+β)=sinαcosβ + cosαsinβ
=4√3/7*cosβ +1/7*sinβ
=5√3/14
解上述关于 sinβ、cosβ 的二元一次方程组,得
sinβ =√3/2、
cosβ = 1/2.