Question

Помогите пожалуйста) 

а)arccos(-1) - 2 arcctg 0=

б)arcsin (-корень из 3/2) + arctg корень из 3=

в)arccos (sin(-п/4))=

г)arcsin(-1) + 2 arctg 0=

д)arccos (-1/2) - 2 arcctg корень из 3=

е)arccos ( tg(-п/4))= 

Answer 1

1) П-2*П/2=0
2) arcsin(-√3/2)+arctg(√3)=-П/3+П/3=0
3)arccos(sin(-п/4))=arccos(-√2/2)=3П/4
4)-П/2+2*0=-П/2
5)2П/3-2П/3=0
6)arccos(tg(-П/4))=arccos(-1)=П

Answer 2

а)arccos(-1) - 2 arcctg 0 = π-2·(π/2) = π-π = 0; б)arcsin (-√3/2) + arctg √ 3 = (-π/3)+(π/3) = 0; в)arccos (sin(-п/4)) = arccos(-√2/2) = π-arccos(√2/2) = π-(π/4) = 3π/4;

г)arcsin(-1) + 2 arctg 0 = (-π/2)+2·0 = (-π/2);

д)arccos (-1/2) - 2 arcctg √3= (π-arccos(1/2))-2·(π/6)=(π-(π/3))-(π/3)=π;

е)arccos ( tg(-п/4))=arccos(-1)=π.