Вычислите : 1)2arcsin (-√3/2) + arctg(-1) + arccos√2/2 - вопрос №6359966123212212 от NikitaPetrov1234 17.03.2023 17:16

Question

Математика: Вычислите : 1)2arcsin (-√3/2) + arctg(-1) + arccos√2/2 - arcctg1; 2) arccos(-1) - arcsin0 + arcctg (-√3) + arctg 1/√3

NikitaPetrov1234 · Answer

1)2arcsin (-√3/2) + arctg(-1) + arccos√2/2 - arcctg1= 2*(-π/3) -π/4 + π/4 -π/4 =
= -2π/3 -π/4 = -11π/12
2) arccos(-1) - arcsin0 + arcctg (-√3) + arctg 1/√3 =π - 0 - π/6 + π/6 = π