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albinastyles00
@albinastyles00
November 2021
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Помогите с 1 и 4 пожалуйста
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nafanya2014
Verified answer
1)
cos(-π/3)=cos(π/3)=1/2
sin(π/6)=1/2
tg(π/4)=1
sinπ=0
ctg(π/6)=√3
(4·(1/2)+2·(1/2)-5):(0+2√3)=-3/2√3=-√3/2
4)
tg(π+β)=tgβ
ctg((π/2)-β)=tgβ
ctg((3π/2)-β)=tgβ
tg(π-β)=-tgβ
tg(π+β)·ctg((π/2)-β)-ctg((3π/2)-β)·tg(π-β)=tgβ·tgβ-tgβ·(-tgβ)=2tg²β
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Answers & Comments
Verified answer
1)cos(-π/3)=cos(π/3)=1/2
sin(π/6)=1/2
tg(π/4)=1
sinπ=0
ctg(π/6)=√3
(4·(1/2)+2·(1/2)-5):(0+2√3)=-3/2√3=-√3/2
4)
tg(π+β)=tgβ
ctg((π/2)-β)=tgβ
ctg((3π/2)-β)=tgβ
tg(π-β)=-tgβ
tg(π+β)·ctg((π/2)-β)-ctg((3π/2)-β)·tg(π-β)=tgβ·tgβ-tgβ·(-tgβ)=2tg²β