Arctg1-2arcctg в корне3

Помогите

Sin (arctg (1/2) - arcctg (-√3)) = sin (arcsin (1/√3) + π-arcctg (√3)) = - sin (arcsin (1/√3) + arctg (1/√3)) =

=-sin (arcsin (1/√3) + arcsin (1/2)) = - sin (arcsin (1/√3)) cos (arcsin (1/2)) - sin (arcsin (1/2) cos (arcsin (1/√3)) = - (1/√3) · (√3/2) - (1/2) · (√2/√3) = - 1/2-1/√6.