Помогите решить 1)cos (arcsin 1/2)-arccos 1 2) cos(3arctg корень 3-arccos(-1/2))
cos(arcsin 1/2)=cosp/6=k3/2 cos(arcsin 1/2)-arccos1=cosp/6-arccos1=k3/2-2p
cos(3arctgk3-arccos(-1/2))=cos(p-2p/3)=cos2p/3=-1/2
Оцени ответ
cos(arcsin 1/2)=cosp/6=k3/2 cos(arcsin 1/2)-arccos1=cosp/6-arccos1=k3/2-2p
cos(3arctgk3-arccos(-1/2))=cos(p-2p/3)=cos2p/3=-1/2