Question

Объясните как решать!!!!!
Arccos1/2-arccos√3/2
Sin(Arccos(-1/2))
Tg(arccos√3/2)

Answer 1

Решение
Arccos1/2-arccos√3/2 = π/3 - π/6 = π/3
Sin(Arccos(-1/2)) = sin(2π/3) = sin(3π - π)3 = sin(π - π/3) = sinπ/3 = √3/2
Tg(arccos√3/2) = tg(π/6) = 1/√3