Помогите 1 и 2 задание

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