已知向量a=(sin(α+π/6),1),b=(4,4cosα-根号3),若a⊥b,则sin(α+4π/3)等于

sin(α+π/6)=sinα·cos(π/6)+cosα·sin(π/6)=(√3/2)sinα+(1/2)cosα;
所以sin(α+π/6)+cosα=(√3/2)sinα+(3/2)cosα;
即(√3/2)sinα+(3/2)cosα=√3/4
sinα+(√3)cosα=1/2;
(1/2)sinα+(√3/2)cosα=1/4
即sinα·cos(π/3)+cosα·sin(π/3)=1/4;
即sin(α+π/3)=1/4