вправа 13.26 гдз 10 клас математика Істер 2018

 
Вправа 13.26


Доведіть тотожність:
1) cos(α - β) - cos(π/2 - α)cos(π/2 - β) = cosαcosβ;
2) (cos(α-β)) / sinαsinβ = ctg2ctgβ + 1;
3) sin(π/4 + β) = cos(π/4 - β);
4) (cos(α-β)-2sinαsinβ) / (2sinαcosβ-sin(α-β)) = ctg(α + β).

 

Умова:


Відповідь - ГДЗ:

1) cos(α - β) - cos(π/2 - α)cos(π/2 - β) = cosαcosβ
cos(α - β) - cos(π/2 - α)cos(π/2 + β) =
= cosαcosβ + sinαsinβ - sinα • sinβ = cosαcosβ;
2) (cos(α-β)) / sinαsinβ = ctg2ctgβ + 1
(cosαcosβ+sinαsinβ) / (sinαsinβ) =
= (cosαcosβ) / (sinαsinβ) + (sinαsinβ) / (sinαsinβ) =
= ctgαctgβ + 1;
3) sin(π/4 + β) = cos(π/4 - β)
sin π/4 cosβ + cos π/4 sinβ = √2/2 cosβ + 2/2 sinβ =
= cos π/4 cosβ + sin π/4 sinβ = cos(π/4 - β);
4) (cos(α-β)-2sinαsinβ) / (2sinαcosβ-sin(α-β)) = ctg(α + β)
(cos(α-β)-2sinαsinβ) / (2sinαcosβ-sin(α-β)) =
= (cosαcosβ+sinαsinβ-2sinαsinβ) / (2sinαаcosβ-sinαcosβ+cosαsinβ) =
= (cosαcosβ-sinαsinβ) / (cosαsinβ+sinαcosβ) =
= (cos(α+β)) / (sin(α+β)) = ctg(α + β).