Ответы
https://znanija.com/task/34258866
* * * * * * * * * * * * * * * * * * * * * * * * * *
4. Докажите:
1) cos(45° -x)*cosx -sin(45°-x)*sinx = 1/√2
* * * имеем cosα*cosβ -sinα*sinβ = cos(α + β) * * *
cos(45° -x)*cosx -sin(45°-x)*sinx =cos(45°-x+x) =cos45°= 1/√2 .
2) ( cosα*sin(α-3) - sinα*cos(3 -α) ) / (cos(3 -π/6) -0,5sin3) = -2√3 /3tg3
* * * имеем sinα*cosβ -cosα*sinβ =sin(α - β) * * *
( cosα*sin(α-3) - sinα*cos(3 -α) ) / (cos(3 -π/6) -0,5sin3) =
( sin(α-3)*cosα- cos(α -3)*sinα ) / (cos3*cosπ/6+sin3*sinπ/6 -0,5sin3) =
sin(α-3-α) / (cos3*√3 /2+sin3*1/2-0.5sin3) =sin(-3) / √3 /2cos3=
- sin3/cos3 * 2/√3= - (2√3 /3) tg3
3) ( sinα+cosα) /cosα*sinα(tgα+ctgα) - 1/√(1+tg²α) = sinα
при -π/2 <α<π/2
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* * * известно sin²α+cos²α = 1 и 1+tg²α = 1/cos²α * * *
(sinα+cosα) /cosα*sinα(tgα+ctgα) - 1/√(1+tg²α) =
(sinα+cosα) /( cosα*sinα*tgα+cosα*sinα*ctgα) ) - 1/√(1+sin²α/cos²α)=
(sinα+cosα) /(sin²α+cos²α) - 1/√( 1/cos²α) = sinα+cosα - √cos²α=
sinα+cosα - |cosα|= sinα+cosα - cosα = sinα
т.к. при -π/2 <α<π/2 cosα > 0 ⇒|cosα| =cosα
5. Решите уравнение :
1) sin(x+π/3)*cos2x - cos(x+π/3)*sin2x =0,5
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* * * sinα*cosβ -cosα*sinβ =sin(α - β) * * *
sin(x+π/3-2x) - cos(x+π/3)*sin2x =0,5
sin(π/3-x) =0,5 ⇔ - sin(x -π/3) = 0,5⇔ sin(x -π/3) = - 0,5⇒
x -π/3 = ( -1) ⁿ⁺¹ π/6 +πn , n∈ℤ
ответ : x=π/3+( -1) ⁿ⁺¹ π/6 +πn , n ∈ ℤ .
ИЛИ иначе
x₁ = π/6 +2πk , при четных n n =2k ∈ℤ ;
x₂ = 3π/2 +2πk , при нечетных n=2k+1 ∈ℤ n∈ℤ