Question

arctg1-2arcctg в корне3
Помогите

Answer 1

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.