Question

Дано tg a=3/2 (0 Найдите sin2a, cos2a,tg2a

Answer 1

Решение
tg a=3/2 ( Найдите sin2a, cos2a,tg2a)
sin2a = (2tga)/(1 + tg²a) = (2*1,5) / (1 + 1,5²) =
 = 3/(3,25) = 300/325 = 12/13
cos2a = (1 - tg²a)/(1 + tg²a) = (1 - 1,5²)/( 1 + 1,5²) =
= (- 1,25) / (3,25) =  - 5 / 13
tg2a = sin2a : cos2a = 12/13 : (- 5/13) = - 12/5

Answer 2

Спасибо)