Question

sin α= 5/13 , π/2<α<π Найти sin 2α, cos 2α, tg 2α. C объяснением

Answer 1

Ответ:

cos^2 (a) + sin^2(a) = 1

cos^2(a) = 169/169 - 25/169 = 144/169

cos(a) = - 12/13 ( т. к. 3я четверть)

sin(2a) = 2sin(a) * cos(a) = 2 * 12/13 * 5/13 = 120/169