B6. sin α=√3/2 ⇒α=π/3
sin(π/6+π/3)= sinπ/2=1
B7. cosα=-1/2 ⇒ cos2α=2cos^2α-1=2(-1/2)^2-1=-1/2 sinα=√(1-(-1/2)^2)=√3/4=√3/2
sin2α=2sinα·cosα=2·(√3/2)·(-1/2)=-√3/2
cos(π/2+α)=-sinα=-√3/2
тогда получим: (1-1/2+√3/2)/(-1/2-√3/2)=-(1/2+√3/2)/(1/2+√3/2)=-1