Question

Помогите 1 и 2 задание

Answer 1

1) sin a = -5/13
3π/2 < a  < 2π  ⇒ cos a > 0
sin² a + cos² a = 1
$cos{a}=+sqrt{1-sin^2{a}}=sqrt{1-(frac{-5}{13})^2}=sqrt{frac{144}{169}}=frac{12}{13}\tg{a}=frac{sin{a}}{cos{a}}=frac{-5}{12}\ctg{a}=frac1{tg{a}}=frac{-12}{5}=-2,4$

2)  cos 1290° = cos (3*360 + 210) = cos 210° = cos (180 + 30) = -cos 30° = -0,5√3

cos 15π/4 = cos [(16π - π) /4] = cos(4π - π/4) = cos (2* 2π - π/4) = cos (-π/4) = cos π/4 = 0,5√2 

Answer 2

Спасибо)