Пусть sina=5/13, a∈ ii четверти найти sin2a, cos2a,tg2a

Sinα=5/13;(|| четверть⇒90°<α<180°)
sin²α+cos²α=1;⇒
cos²α=1-sin²α=1-(5/13)²=144/169;
cosα=-12/13;
sin2α=2sinαcosα;
sin2α=2·(5/13)·(-12/13)=-120/169;
cos2α=cos²α-sin²α;
cos2α=144/169-25/169=119/169;
tg2α=sin2α/cos2α=(-120/169):(119/169)=-12/119;