1. 2)
2. 3)
Объяснение:
1.
2.
верный ответ: tg(90°+40°)=−ctg40°
1) (sin(α−β)+2cosα⋅sinβ)/(2cosα⋅cosβ−cos(α−β))=
=(sinαcosβ-cosαsinβ+2cosα⋅sinβ)/(2cosα⋅cosβ−cosαcosβ-sinαsinβ)=
=(sinαcosβ+cosα⋅sinβ)/(cosα⋅cosβ-sinαsinβ)=sin(α+β)/cos(α+β)
2) ctg130°=ctg(90°+40°)= - tg40°
sin(π + 10) = - sin10°
cos(π + 10) = - cos10°
tg(π + 10) = tg10°
ctg(π + 10) = ctg10°
sin(190)=sin(180+10)=-sin10 (sin(190)<0)
cos(190)=cos(180+10)=-cos10 (cos(190)<0)
tg(190)=tg(180+10)=tg10 (tg(190)>0)
ctg(190)=ctg(180+10)=ctg10 (ctg(190)>0)
1. 2)
2. 3)
Объяснение:
1.
2.
верный ответ: tg(90°+40°)=−ctg40°
Объяснение:
1) (sin(α−β)+2cosα⋅sinβ)/(2cosα⋅cosβ−cos(α−β))=
=(sinαcosβ-cosαsinβ+2cosα⋅sinβ)/(2cosα⋅cosβ−cosαcosβ-sinαsinβ)=
=(sinαcosβ+cosα⋅sinβ)/(cosα⋅cosβ-sinαsinβ)=sin(α+β)/cos(α+β)
2) ctg130°=ctg(90°+40°)= - tg40°
cos130=cos(90+40)=-sin40
tg130=tg(90+40)=-ctg40
ctg130=ctg(90+40)=-tg40
или
sin130=sin(180-50)=sin50
cos130=cos(180-50)=-cos50
tg130=tg(180-50)=-tg40
ctg130=ctg(180-50)=-ctg40
sin(130)=sin(180-50)=sin 50
sinC=sin(π/2-α);