ВНИМАНИЕ! Спасибо за внимание, пожалуйста, помогите вычислить)
1)=cos(1/2*π/3-3*π/2-2π/3)=cos(π/6-3π/2-2π/3)=cos(-2π)=1 2)=1/3(arccos1/3+π-arccos1/3)=π/3
1) 2arccos(1/2)=2*(π/3)=2π/3; 3arccos0=3π/2; arccos(-1/2)=π-arccos(1/2)=π-π/3=2π/3 cos(2arccos(1/2)-3arccos0-arccos(-1/2))=cos(2π/3-3π/2-2π/3)= cos(-3π/2)=cos(3π/2)=0 2) 1/3(arccos(1/3)+arccos(-1/3))=1/3(arccos(1/3)+π-arccos(1/3))=π/3