a){y+2x=-5
{x^2+y^2=25
{y=-5-2x
{x^2+(-5-2x)^2=25
x^2+4x^2+10x+25=25
5x^2+10x=0
5(x^2+2x)=0
x^2+2x=0
x(x+2)=0
x=0
x+2=0
x=-2
y=-5-2*-2=-9
y=-5-2*0=-5
ответ (0 -2) (-9 -5)
б)
{y=x^2+6x+7
{y-2x=4
y=4+2x
y=x^2+6x+7
4+2x=x^2+6x+7
x^2+4x+3=0
D=16-4*1*3=V4=2
x=-4+2/2=-1
x2=-6/2=-3
y=4+2*-1=-2
y2=4+2*-3=-2
ответ (-1 -3) (-2 -2)
в)
{xy-2y-4x=-5
{x-3y=-10
x=-10+3y
(-10+3y)y-2y-4(-10+3y)=-5
-10y+3y^2-2y+40-12y=-5
3y^2-24y+45=0
D=4
y=3
y=5
x=-10+3*3=-1
x=-10+3*5 =5
ответ (3 5 ) ( -1 5)
(sin150°)/2 =(sin(180°- 30°))/2 = (sin30°)/2 =(1/2) /2 =1/4.
1) sin105°*sin75° = sin(180° -75°)*sin75° = sin75°*sin75° =sin²75°=
(1 -cos2*75°)/2 =(1 -cos150°)/2 = (1 -cos(180° -30°) )/2 = (1+cos30°) /2 =
(2+√3) / 4 .
* * * sin²75° =(sin45°cos30° + cos45°sin30°) ² = ( (1/√2)*(√3)/2 +(1/√2)*(1)/2) ) ² =(1/8) *(√3 +1) ² =(1/8) *(3 +2√3 +1)= (1/4) *(2 +√3 )= (2 +√3 ) /4.
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2) 4sin(π/6 -β)cos(π/6+β)= 4 *(sin(π/6 -β+π/6+β) + sin(π/6 -β-π/6-β) )/2 =
2 *(sin π/3 + sin( -2β) ) = 2 *( (√3)/2 - sin2β ) =√3 -2 sin2β.
* * * А если преобразование начнем с правой стороны равенства , то
3 - 4cos²β = 4(1 - cos²β) -1 =4sin²β -1 =2*2sin²β -1 =2(1 -cos2β) -1 =
2(1 - cos2β -1/2) = 2(1/2 -cos2β) = 2(cosπ/3 -cos2β) = 2(cosπ/3 -cos2β) =
- 4sin(π/6- β)*sin(π/6+ β) .