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lark1972
@lark1972
July 2022
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Помогите! Номер 17!
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oganesbagoyan
Verified answer
17.
а)
=(cos²α-sin²α)/(cosα+sinα)² =(cosα-sinα)(cosα+sinα)/(cosα+sinα)² =
=(cosα-sinα)/(cosα+sinα) =
cosα
(1-tqα)/
cosα
(1+tqα) =(1-tqα)/(1+tqα).
-------
б)
= 2cos²2α/(cosα/sinα -sinα/cosα) =2cos²2α*sinαcosα/(cos²α-sin²α)=
cos²2α*sin2α/cos2α =cos2α*sin2α= (1/2)*sin4α.
-------
в)
= (tqα +2√(tqα*ctqα) +ctqα)/(sinα+cosα)²=
(sinα/cosα +2 +cosα/sinα)/(sinα+cosα)² =
(sin²α+2sinαcosα+cos²α)/sinαcosα(sinα+cosα)²=
2(sinα+cosα)²/2sinαcosα(sinα+cosα)² = 2/sin2α.
-------
г)
= (sin2α+2sinαcos4α)/(cosα+cos4α) =
(2sinα*cosα+2sinαcos4α)/(cosα+cos4α) =2sinα(cosα+cos4α)/cosα+cos4α) =2sinα.
* * * * * * *
удачи
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Answers & Comments
Verified answer
17.а) =(cos²α-sin²α)/(cosα+sinα)² =(cosα-sinα)(cosα+sinα)/(cosα+sinα)² =
=(cosα-sinα)/(cosα+sinα) = cosα(1-tqα)/cosα(1+tqα) =(1-tqα)/(1+tqα).
-------
б) = 2cos²2α/(cosα/sinα -sinα/cosα) =2cos²2α*sinαcosα/(cos²α-sin²α)=
cos²2α*sin2α/cos2α =cos2α*sin2α= (1/2)*sin4α.
-------
в) = (tqα +2√(tqα*ctqα) +ctqα)/(sinα+cosα)²=
(sinα/cosα +2 +cosα/sinα)/(sinα+cosα)² =
(sin²α+2sinαcosα+cos²α)/sinαcosα(sinα+cosα)²=
2(sinα+cosα)²/2sinαcosα(sinα+cosα)² = 2/sin2α.
-------
г) = (sin2α+2sinαcos4α)/(cosα+cos4α) =
(2sinα*cosα+2sinαcos4α)/(cosα+cos4α) =2sinα(cosα+cos4α)/cosα+cos4α) =2sinα.
* * * * * * *
удачи