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Juliya11
@Juliya11
August 2022
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Решите, пожалуйста. Хочу себя проверить. Алгебра 10 класс
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oganesbagoyan
Verified answer
* * * * * * * cos(-β) =cosβ ; sin(-β) = - sinβ ; tq(-β) = - tqβ ; ctq( -β) = - ctqβ * * * * * * *
а)
(1+cos(-β))/sin(-β) - ctq(-β) = (1+cosβ)/ (-sinβ) +ctqβ = - (1+ (-1-cosβ+sinβ*ctqβ )/sinβ =cosβ)/sinβ +ctqβ= (-1-cosβ+sinβ*ctqβ)/sinβ = (-1-cosβ+cosβ)/sinβ =
-1/sinβ
.
-------------
Можно сразу заменить (-β) =α :
(1+cosα)/sinα -ctqα = (1+cosα -sinα*ctqα)/sinα = (1+cosα -cosα)/sinα =1/sinα →
=1/sin(-β) = 1/(-sinβ) = - 1/sinβ .
------------
б)
sin²(-β) - sin⁴(-β)
)
/ cos²( -β) = (sin² β -sin⁴β)/ cos²β =sin²β(1-sin²β)/ cos²β =
sin²β*cos²β / cos²β = sin²β .
------------
в)
sin(
-
α)/(1-cos(-α) ) -ctq(-α) = -
sinα/(1-cosα) +ctqα = sinα/(cosα -1) +sinα/cosα =
sinα*cosα+sinα*cosα -sinα)/cosα(cosα -1) = (2sinαcosα -sinα)/cosα(cosα -1) =
sinα(2cosα -1)/cosα(cosα -1)
(
-sinα +ctqα(1- cosα)
)
/(1-cosα)
=( -sinα +ctqα - ctqα *cosα)/(1-cosα) = =( -sinα +ctqα - sinα)/(1-cosα)
2 votes
Thanks 1
Juliya11
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Answers & Comments
Verified answer
* * * * * * * cos(-β) =cosβ ; sin(-β) = - sinβ ; tq(-β) = - tqβ ; ctq( -β) = - ctqβ * * * * * * *а) (1+cos(-β))/sin(-β) - ctq(-β) = (1+cosβ)/ (-sinβ) +ctqβ = - (1+ (-1-cosβ+sinβ*ctqβ )/sinβ =cosβ)/sinβ +ctqβ= (-1-cosβ+sinβ*ctqβ)/sinβ = (-1-cosβ+cosβ)/sinβ = -1/sinβ.
-------------
Можно сразу заменить (-β) =α :
(1+cosα)/sinα -ctqα = (1+cosα -sinα*ctqα)/sinα = (1+cosα -cosα)/sinα =1/sinα →
=1/sin(-β) = 1/(-sinβ) = - 1/sinβ .
------------
б) sin²(-β) - sin⁴(-β) ) / cos²( -β) = (sin² β -sin⁴β)/ cos²β =sin²β(1-sin²β)/ cos²β =
sin²β*cos²β / cos²β = sin²β .
------------
в) sin(-α)/(1-cos(-α) ) -ctq(-α) = - sinα/(1-cosα) +ctqα = sinα/(cosα -1) +sinα/cosα =
sinα*cosα+sinα*cosα -sinα)/cosα(cosα -1) = (2sinαcosα -sinα)/cosα(cosα -1) =
sinα(2cosα -1)/cosα(cosα -1)
(-sinα +ctqα(1- cosα)) /(1-cosα)
=( -sinα +ctqα - ctqα *cosα)/(1-cosα) = =( -sinα +ctqα - sinα)/(1-cosα)