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stiklin
@stiklin
July 2022
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Номер 44(1) и 45(2)
Задание во вложении !!)))
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Task/25017191
-------------------
44 (1) Упросить выражение
( sin²(3π/2 - α)+cos²(3π - α) +cos(π+α)*cos(2π-α) ) /
(tg²(α-π/2)*ctg²(3π/2 +α) =( cos²α+cos²α - cosα*cosα) / (ctg²α*tq²α) =
cos²α /(ctgα*tqα)² =cos²α /1²
= cos²α .
------------------
45 (2) Найдите значение выражения (sinα+3cosα) / (sin³α - 3cos³α) ,
если tgα =3 :
-----------
Т.к. существует tgα , значит cosα ≠ 0 .
(sinα+3cosα) / (sin³α - 3cos³α) =cosα(sinα/cosα+3) / cos³α(sin³α/cos³α - 3) =
(1/cos²α)*(tgα+3) / (tg³α - 3) =(1+tg²α)(tgα+3) /(tg³α - 3) = || если tgα =3 || =
(1+3²)(3+3) /(3³ -3) = 60/24 =5/2
= 2,5 .
2 votes
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stiklin
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Verified answer
Task/25017191-------------------
44 (1) Упросить выражение
( sin²(3π/2 - α)+cos²(3π - α) +cos(π+α)*cos(2π-α) ) /
(tg²(α-π/2)*ctg²(3π/2 +α) =( cos²α+cos²α - cosα*cosα) / (ctg²α*tq²α) =
cos²α /(ctgα*tqα)² =cos²α /1² = cos²α .
------------------
45 (2) Найдите значение выражения (sinα+3cosα) / (sin³α - 3cos³α) ,
если tgα =3 :
-----------
Т.к. существует tgα , значит cosα ≠ 0 .
(sinα+3cosα) / (sin³α - 3cos³α) =cosα(sinα/cosα+3) / cos³α(sin³α/cos³α - 3) =
(1/cos²α)*(tgα+3) / (tg³α - 3) =(1+tg²α)(tgα+3) /(tg³α - 3) = || если tgα =3 || =
(1+3²)(3+3) /(3³ -3) = 60/24 =5/2 = 2,5 .