求值arcsin(-根号3/2)+arctg(-1)+2arccos(-根号3/2
问题描述：
求值arcsin(-根号3/2)+arctg(-1)+2arccos(-根号3/2
答
arcsin(-sqrt(3)/2)=-pi/3;
arctg(-1)=-pi/4+pi=(3/4)pi;
arccos(-sqrt(3)/2)=(5/6)pi;
then,answer=(25/12)pi
pay attention to the dingyiyu