Question

помогите пожалуйста решить даю 19 баллов

Answer 1

1)sin(α-β)-sin(π/2-α)sin(-β)=
=sinαcosβ-cosαsinβ-cosα·(-sinβ)=
=sinαcosβ-cosαsinβ+cosαsinβ=sinαcosβ;
2)cos²(π-α)-cos²(π/2-α)=(-cosα)²-sin²α=cos²α-sin²α=cos2α;
3)2sinαsinβ+cos(α+β)=2sinαsinβ+cosαcosβ-sinαsinβ=
=sinαsinβ+cosαcosβ=
=1/2[cos(α-β)-cos(α+β)]+1/2[cos(α-β)+cos(α+β)]=
=cos(α-β);