已知π/2

问题描述:

已知π/2

π/2π0cos( α-β)=12/13==> sin( α-β)=5/13,
sin(α+β)=-3/5==>cos(α+β)=-4/5
sin2α
=sin(α+β + α-β)
=sin(α+β)cos(α-β)+cos(α+β)sin(α-β)
=-3/5*12/13 -4/5*5/13
=-56/65
cos2β
=cos(α+β - α-β)
=cos(α+β)cos(α-β)+sin(α+β)sin(α-β)
=-4/5*12/13 -3/5*5/13
=-63/65
=