Ответы
Ответ дал:
0
a)sin²(π/3+β)+sin²(π/3-β)+sin²β=[1-cos(2π/3+2β)]/2+[1-cos(2π/3-2β)]/2+(1-cos2β)/2=[3-(cos(2π/3+2β)+cos(2π/3-2β)+cos2β)]/2=
[3-(2cos2π/3cos2β+cos2β)]/2=[3-cos2β(2cos2π/3+1)]/2=[3-cos2β(2·(-1/2)+1)]/2=
[3-cos2β(-1+1)]/2=3/2
г)ctgαctgβ+1=cosαcosβ/sinαsinβ+1=(cosαcosβ+sinαsinβ)/sinαsinβ=
cos(α-β)/sinαsinβ
1-ctgαctgβ=1-cosαcosβ/sinαsinβ=(sinαsinβ-cosαcosβ)/sinαsinβ=
-(cosαcosβ-sinαsinβ)/sinαsinβ=-cos(α+β)/sinαsinβ
cos(α+β)·cos(α-β)/sinαsinβ-cos(α+β)cos(α-β)/sinαsinβ=0
[3-(2cos2π/3cos2β+cos2β)]/2=[3-cos2β(2cos2π/3+1)]/2=[3-cos2β(2·(-1/2)+1)]/2=
[3-cos2β(-1+1)]/2=3/2
г)ctgαctgβ+1=cosαcosβ/sinαsinβ+1=(cosαcosβ+sinαsinβ)/sinαsinβ=
cos(α-β)/sinαsinβ
1-ctgαctgβ=1-cosαcosβ/sinαsinβ=(sinαsinβ-cosαcosβ)/sinαsinβ=
-(cosαcosβ-sinαsinβ)/sinαsinβ=-cos(α+β)/sinαsinβ
cos(α+β)·cos(α-β)/sinαsinβ-cos(α+β)cos(α-β)/sinαsinβ=0
Вас заинтересует
1 год назад
2 года назад
8 лет назад
8 лет назад