Question

Arccos(sin27п/7) помогите пожалуйста

Answer 1

arccoc(числа) = угол (0 ≤ угол ≤ π)

sin(27п/7) = sin(14п/7 + 13п/7) = sin(2п + 13п/7) = sin(13п/7) =
= sin(26п/14) = sin(21п/14 + 5п/14) = sin(3п/2 + 5п/14) = -cos(5п/14) 

arccos(-x) = п - arccos(x)           arccos(cos(y)) = y, при 0 ≤ y ≤ п

arccos(sin(27п/7)) = arccos(-cos(5п/14)) = п - arccos(cos(5п/14)) = 
= п - 5п/14 = (14 - 5)п/14 = 9п/14