Question

Найдите остальные тригонометрические функции
Sin x = 1213
0 < x < π2

Answer 1

Решение
Sin x = 1213
0 < x < π2
cosx = √(1 - sin²x) = √(1 - (12/13)²) = √(1 - 144/169) =
= √25/169 = 5/13
tgx = sinx/cosx = 12/13 : 5/13 = 12/5
ctgx = 1/tgx = 5/12