Question

Помогите,пожалуйста!

Answer 1

Відповідь:

a) sin(2/3π) = √3/2

  cos(2/3π) = - 1/2

  tg(2/3π) = -√3

  ctg(2/3π) = -√3/3

б)sin(3/4π) = √2/2

  cos(3/4π) = - √2/2

  tg(3/4π) = -1

  ctg(3/4π)= -1

в)sin(5/6π) = 1/2

  cos(5/6π)  = -√3/2

  tg(5/6π)  = -√3/3

  ctg(5/6π)  = -√3