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yanaberezhnaya
@yanaberezhnaya
July 2022
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1) tg(-a)*ctga+sin^2(-a) если tga=-3/4
2) (sina*cosa)/(sin^2a-cos^2a) если tg=3/2
3) корень из 5 *sina если tg=2 a€3четверти
4) tg^3a+ctg^3a если tga + ctga=5
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oganesbagoyan
Verified answer
task/27281836
-------------------
1)
tg(-
α)*ctgα+sin²(-α
) , если tg
α
=-3/4
---
tg(-α)*ctgα+sin²(-α) = -tgα*ctgα+sin²α= -1 + sin²α = -(1-sin²α) = - cos²α =
- ( 1/(1+tg²α) = -1/(1 +(-3/4)² )
= -16/25 .
-------
2)
(sinα*cosα)/(sin^2α-cos^2α) , если tgα=3/2
---
(sinα*cosα)/(sin²α-cos²α) = (1/2)*sin2α / (- cos2α) = -(1/2)*tg2α =
- tgα / (1+tg²α) = -(3/2) /(1 +(3/2)²)
= - 6/13 .
-------
3)
√5 *sina , если tgα=2 ,α ∈
3 четверти
---
α ∈ 3 четверти ⇒ sinα < 0
√5 *sinα =√5 *(- √(1 -cos²α) )= - √5 *√(1 - 1/(1+tg²α) ) = - √5 *√(1 - 1/(1+2²) )=
- √5 *√(1 - 1/5) = - √5 *√(4/5) = - √5 *2/√5
= - 2.
-------
4)
tg^3a+ctg^3a , если tga + ctga=5
---
tg³α+ctg³α= (tgα+ctgα)³ -3tgα*ctgα(tgα+ctgα) = (tgα+ctgα)³ -3*1*(tgα+ctgα)=
5³ -3*5 =125 -15
=110.
1 votes
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Answers & Comments
Verified answer
task/27281836-------------------
1)
tg(-α)*ctgα+sin²(-α) , если tgα=-3/4
---
tg(-α)*ctgα+sin²(-α) = -tgα*ctgα+sin²α= -1 + sin²α = -(1-sin²α) = - cos²α =
- ( 1/(1+tg²α) = -1/(1 +(-3/4)² ) = -16/25 .
-------
2)
(sinα*cosα)/(sin^2α-cos^2α) , если tgα=3/2
---
(sinα*cosα)/(sin²α-cos²α) = (1/2)*sin2α / (- cos2α) = -(1/2)*tg2α =
- tgα / (1+tg²α) = -(3/2) /(1 +(3/2)²) = - 6/13 .
-------
3)
√5 *sina , если tgα=2 ,α ∈ 3 четверти
---
α ∈ 3 четверти ⇒ sinα < 0
√5 *sinα =√5 *(- √(1 -cos²α) )= - √5 *√(1 - 1/(1+tg²α) ) = - √5 *√(1 - 1/(1+2²) )=
- √5 *√(1 - 1/5) = - √5 *√(4/5) = - √5 *2/√5 = - 2.
-------
4)
tg^3a+ctg^3a , если tga + ctga=5
---
tg³α+ctg³α= (tgα+ctgα)³ -3tgα*ctgα(tgα+ctgα) = (tgα+ctgα)³ -3*1*(tgα+ctgα)=
5³ -3*5 =125 -15 =110.