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228akmal228
@228akmal228
July 2022
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Решите пожалуйста тригономертию
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ВладимирБ
Verified answer
Решение смотри на фото
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oganesbagoyan
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Доказать тождество
=================
sin2α*(1-tq²α)/1+tq²α) ==sin4α.
---
2sin2α*(1-tq²α)/1+tq²α) =2sin2α*(cos²α- sin²α)/(sin²α+cos²α) =
2sin2α*cos2α =sin4α.
-----
( 4sinα*cosα*cos2α/(cos²2α - sin²2α) )*ctq4α =1.
---
(
2sinα*cosα*cos2α / (cos²2α - sin²2α)
)
*ctq4α =
(
2sin2α*cos2α / cos(2*2α)
)
*ctq4α =sin(2*2α)/cos4α =(sin4α /cos4α )*ctq4α =tq4α*ctq4α =1. * * * * * * * * P.S.
sin2β =2sinβ*cosβ ; cos2β =cos²β -sin²β ;tqβ =sinβ/cosβ ;ctqβ =cosβ/sinβ.
+ + + 50 + + + sin²α+cos²α =1.
0 votes
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Answers & Comments
Verified answer
Решение смотри на фотоVerified answer
Доказать тождество=================
sin2α*(1-tq²α)/1+tq²α) ==sin4α.
---
2sin2α*(1-tq²α)/1+tq²α) =2sin2α*(cos²α- sin²α)/(sin²α+cos²α) =
2sin2α*cos2α =sin4α.
-----
( 4sinα*cosα*cos2α/(cos²2α - sin²2α) )*ctq4α =1.
---
( 2sinα*cosα*cos2α / (cos²2α - sin²2α) )*ctq4α =
( 2sin2α*cos2α / cos(2*2α) )*ctq4α =sin(2*2α)/cos4α =(sin4α /cos4α )*ctq4α =tq4α*ctq4α =1. * * * * * * * * P.S.
sin2β =2sinβ*cosβ ; cos2β =cos²β -sin²β ;tqβ =sinβ/cosβ ;ctqβ =cosβ/sinβ.
+ + + 50 + + + sin²α+cos²α =1.