Question

Помогите срочно решить

Answer 1

235.

a) arctg(1)-arctg(√3)--arccos(-0,5)=45°-60°-120°=-135°.

б) arcsin(-0,5)+arctg(-1/√3)-arcctg(√3)=-30°+arctg(-√3/3)-30°=

=-60°-30°=-90°.

236.

a) tg(arccos(1)-2*arctg(1/√3))=tg(0°-2*arctg(√3/3)=

=tg(-2*30°)=tg(-60°)=-√3.

б) cos(arctg(3/√3)+arcsin(-√3/2)=cos(arctg(√3)+(-60°))=

=cos(60°-60°)=cos0°=1.

237.

a) sin(2*arcsin(√2/2)+arctg(-√3)=sin(2*45°-60°)=

=sin(90°-60°)=sin30°=0,5.

б) tg(2*arctg(1)+3*arctg(0)+arcsin(-1))=tg(2*45°+3*0+(-90°))= =tg(90°-90°)=tg0°=0.