Question

已知sinα+cosα=1/5,且π/2≤α≤3π/4,求cos2α？

Answer 1

∵sinα+cosα=1/5 ==>(sinα+cosα)²=1/25
==>sin²α+cos²α+2sinαcosα=1/25
==>1+sin(2α)=1/25
∴sin(2α)=-24/25
∵π/2≤α≤3π/4 ==>π≤2α≤3π/2
∴cos(2α),2,(sinα+cosα)²=sin²α+cos²α+2sinαcosα=1+sin2α=1/25，sin2α=—24/25, sin²2α+cos²2α=1 故cos2α=√￣（1-sin²2α）=7/25,1,sinα+cosα=1/5,且π/2≤α≤3π/4,
所以，π/2 - cosa
sinα+cosα=1/5,cosa * cosa +sina * sina+2sina * cosa=1/25;
2sina * cosa=-24/25
cosa * cosa +sina * sina-2sina * cosa=49/25,sinα-cosα=7/5
cos2α=cosa * cosa -sina * sina=(-7/5)*(1/5)=-7/25,0,