Преобразуйте в произведение выражение: а)cos^2a-cos^2b б)3/4-sin^2x в)cos^2x-1/2 г)sin^2a-cos^2п/3 заранее )
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Ответы на вопрос:
А)cos²α-cos²β=(cosα-cosβ)(cosα+cosβ)=-2sin(α-β)/2sin(α+β)/2·2cos(α+β)/2cos(α-β)/2=-2sin(α-β)/2cos(α-β)/2·2sin(α+β)/2cos(α+β)/2=-sin(α-β)sin(α+β) б)3/4-sin²x=(√3/2)²-sin²x=(√3/2-sinx)(√3/2+sinx)=(sinπ/3-sinx)(sinπ/3+sinx)=2sin(π/3+x)/2cos(π/3-x)/2·2sin(π/3-x)/2cos(π/3+x)/2=2sin(π/3+x)/2cos(π/3+x)/2·2sin(π/3-x)/2cos(π/3-x)/2=sin(π/3+x)sin(π/3-x) в)cos²x-1/2=cos²x-(√2/2)²=(cosx-√2/2)(cosx+√2/2)=(cosx-cosπ/4)(cosx+cosπ/4)=-2sin(x-π/4)/2sin(x+π/4)/2·2cos(x+π/4)/2cos(x-π/4)/2=-2sin(x-π/4)/2cos(x-π/4)/2·2sin(x+π/4)/2cos(x+π/4)/2=-sin(x-π/4)sin(x+π/4) г)sin²α-cos²π/3=(sinα-cosπ/3)(sinα+cosπ/3)=(sinα-1/2)(sinα+1/2)=(sinα-sinπ/6)(sinα+sinπ/6)=2sin(α-π/6)/2cos(α+π/6)/2·2sin(α+π/6)/2cos(α-π/6)/2=2sin(α-π/6)/2cos(α+π/6)/2·2sin(α+π/6)/2cos(α+π/6)/2=sin(α-π/6)sin(α+π/6)
{ (xy)²+3y=45
{ 5y-2xy=3
замена: xy = t
{ (t)²+3y=45
{ 5y-2t=3 => t= (5y-3) /2 (подставим значение t в первое ур-ние)
((5y-3) /2 )²+3y=45
(5y-3)²/4 +3y=45 | *4
(5y-3)² + 12y= 180
25y² - 30y + 9 + 12y - 180 = 0
25y² - 18y - 171 = 0
d = 324 + 4*25*171 = 324 + 17100 = 17424
√d = 132
y₁= (18+132) /2*25 = 150/50 = 3
y₂= (1818 - 132) /2*25 = -114/50 = -2,28
подставим значение y₁ и y₂ в уравнение 5y-2xy=3:
5*3-2x*3=3
х₁ = 2
5*(-2,28) - 2х*(-2,28) = 3
- 11,4 + 4,56х = 3
4,56х = 14,4
х₂ = 3 3/19
ответ: (2; 3) , (3 3/19; -2,28)
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