Question

кому под силу помогите ПЖ.19. Вычислите: 3arcsin(1/2) + 4arccos(-1/√2) – arctg(-√3);20. Вычислите: 2arcsin(-√3/2) + arctg(-1) + arccos(√2/2);

Answer 1

3arcsin(1/2) + 4arccos(-1/√2) – arctg(-√3)=3*π/6+4*(-3*π/4)-(-π/3)=π/2-3*π+π/3=π*(1/2-3+1/3)=(5/6-3)*π=-13*π/6;

Вычислите: 2arcsin(-√3/2) + arctg(-1) + arccos(√2/2)=2*(-π/6)+(-π/4)+π/4=-π/3-π/4+π/4=-π/3;