Предмет: Алгебра, автор: ChikaTheBest

найти sin2a cos2a tg2a
если COSA=-12/13
5п/2>а>3п
решите пожалуйста.

Ответы

Автор ответа: natalield

α - угол II четверти
sin2α=2cosα*sinα
найдём sin²α= 1 -cos²α=1-144/169=25/169 ⇒sinα=5/13
sin2α=2*(-12/13)*5/13= -120/169
cos2α= cos²α-sin²α=144/169-25/169=119/169
tg2α= sin2α/cos2α =-120/169 : 144/169= -120/144 =-5/6

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