1-4sin^2t*cos^2t/(sint+cost)^2+2sint*cost=1 - вопрос №61907801422221 от hrapatiynickita 29.01.2020 02:41

Question

Алгебра: 1-4sin^2t*cos^2t/(sint+cost)^2+2sint*cost=1

hrapatiynickita · Answer

Во всей вероятности
(1- 4sin²t·cos²t)/(sint+cost)² + 2sint·cost) =1
(1- sin2t)²/(1+2sint·cost) +sin2t = 1
1- sin2t + sin2t + (sin2t)² = 1+sin2t ;
(sin2t)² - sin2t =0
sin2t·(sin2t -1) = 0
a) sin2t =0 ⇒ 2t = πk ⇒ t= π/2 ·k ; k∈Z
b) sin2t =1 ⇒ 2t = 2πm + π/2 ⇒ t=πm +π/4 ; m∈Z