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;