Помогите пожалуйста решить даю 19 баллов

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);