Question

3arccos(-1/2) - 2arcsi корінь3/2 - arcfg( - корінь3)

Answer 1

arccos(-1/2) = π - (π/3) = 2π/3,

arcsin( (√3)/2) = π/3,

arctg(-√3) = -arctg(√3) = -π/3,

3·arccos(-1/2) - 2·arcsin((√3)/2) - arctg(-√3) = 3·(2π/3) - 2·(π/3) - (-π/3) =

= 2π - (2π/3) + (π/3) = 2π - (π/3) = 5π/3