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nastyastasevich
@nastyastasevich
September 2021
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Пожалуйста, помогите! Алгебра-10 класс! очень надо!!!(ЗАДАНИЕ ВО ВЛОЖЕНИИ)
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oganesbagoyan
Verified answer
1.
а)
(1-sin²α)/cos²α -cosα*tqα =cos²α/cos²α -sinα =
1-sinα .
б)
(sin2α +3cos2α)² +(cos2α- 3sin2α)² =sin²2α +6sin2α*3cos2α +9cos²2α +
cos²2α -6cos2α*sin +9sin²2α =10*(sin²2α +cosα²2α) =
10.
в)
(1-2sinxcosx)/(sinx -cosx) +cosα =(sin²α -2sinαcosα+cos²α)/(sinx -cosx) +cosα =
=(sinα -cosα)²/(sinx -cosx) +cosα =sinx -cosx+cosα=sinα.
-----------------
2)
соsα
=-3/5 ; π<α<3π/2.
----------------
sinα -?
sinα <0 ,если π<α<3π/2.
sinα = -√(1-cos²α) =-√(1 -(-3/5)²) =-√(1 -9/25) =
-4/5.
-------------
3)
tqα = -12/5 , 3π/2 <α <2π .
----------------------------
cosα - α?
1+tq²α =1/cos²α⇒cosα = ± 1/√ (1+tq²α)
cosα >0 , если 3π/2 <α <2π ,
поэтому
cosα
= 1/√ (1+tq²α) =1/√(1 +(-12/5)²
=
5/13.
4)
(1-4sin²αcos²α)/(sinα+cosα)² - 2cosα*sin(-α) =
(1-2sinα*cosα)(1+2sinα*cosα)/(sinα+cosα)² + 2cosα*sinα=
(1-2sinα*cosα)(sinα+cosα)²/(sinα+cosα)² + 2cosα*sinα=
1-2sinαcosα+2cosα*sinα = 1.
* * * 1+2sinα*cosα =sin²α+cos²α +2sinα*cosα =(sinα+cosα)² * * *
5)
sinα -cosα =2/3.
-----------------
sinα*cosα -?
sinα -cosα =2/3.
( sinα -cosα)² =(2/3)² ;
sin²α+2sinα*cosα +cos²α =4/9 ;
1+2sinα*cosα =4/9 ; '
sinα*cosα =
-5/18.
---------------------
6) 2cos²α -3sinα =0 , 0<α<π/2.
---------------
sinα -?
2cos²α -3sinα =0 ;
2(1 -sin²α) -3sinα =0 ;
2sin²α +3sinα -2 = 0 ; * * * sinα =t , -1 ≤ t ≤1 * * *
2t² +3t -2 =0 ;
D =3² -4*2(-2) =5² ;√D =5 ;
t₁ = (-3-5)/2*2 = - 2 не решения
t₂ =(-3+5)/2*2 =
1/2
sinα =1/2
удовлетворяет
0<α<π/2 sinα>0.
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Answers & Comments
Verified answer
1.а) (1-sin²α)/cos²α -cosα*tqα =cos²α/cos²α -sinα = 1-sinα .
б)(sin2α +3cos2α)² +(cos2α- 3sin2α)² =sin²2α +6sin2α*3cos2α +9cos²2α +
cos²2α -6cos2α*sin +9sin²2α =10*(sin²2α +cosα²2α) =10.
в) (1-2sinxcosx)/(sinx -cosx) +cosα =(sin²α -2sinαcosα+cos²α)/(sinx -cosx) +cosα =
=(sinα -cosα)²/(sinx -cosx) +cosα =sinx -cosx+cosα=sinα.
-----------------
2) соsα =-3/5 ; π<α<3π/2.
----------------
sinα -?
sinα <0 ,если π<α<3π/2.
sinα = -√(1-cos²α) =-√(1 -(-3/5)²) =-√(1 -9/25) = -4/5.
-------------
3) tqα = -12/5 , 3π/2 <α <2π .
----------------------------
cosα - α?
1+tq²α =1/cos²α⇒cosα = ± 1/√ (1+tq²α)
cosα >0 , если 3π/2 <α <2π ,
поэтому cosα = 1/√ (1+tq²α) =1/√(1 +(-12/5)² = 5/13.
4) (1-4sin²αcos²α)/(sinα+cosα)² - 2cosα*sin(-α) =
(1-2sinα*cosα)(1+2sinα*cosα)/(sinα+cosα)² + 2cosα*sinα=
(1-2sinα*cosα)(sinα+cosα)²/(sinα+cosα)² + 2cosα*sinα=
1-2sinαcosα+2cosα*sinα = 1.
* * * 1+2sinα*cosα =sin²α+cos²α +2sinα*cosα =(sinα+cosα)² * * *
5) sinα -cosα =2/3.
-----------------
sinα*cosα -?
sinα -cosα =2/3.
( sinα -cosα)² =(2/3)² ;
sin²α+2sinα*cosα +cos²α =4/9 ;
1+2sinα*cosα =4/9 ; '
sinα*cosα = -5/18.
---------------------
6) 2cos²α -3sinα =0 , 0<α<π/2.
---------------
sinα -?
2cos²α -3sinα =0 ;
2(1 -sin²α) -3sinα =0 ;
2sin²α +3sinα -2 = 0 ; * * * sinα =t , -1 ≤ t ≤1 * * *
2t² +3t -2 =0 ;
D =3² -4*2(-2) =5² ;√D =5 ;
t₁ = (-3-5)/2*2 = - 2 не решения
t₂ =(-3+5)/2*2 = 1/2
sinα =1/2 удовлетворяет 0<α<π/2 sinα>0.