Объяснение: * * * cos(-α) =cosα , sin2α=2sinα*cosα и формулы
приведения * * *
1) ( 1 +sin(4π -(π/2 +α) )+ cos(2π-2α) ) / (2sinα*cosα - sinα) =
( 1 - cosα+ cos2α ) / (2sinα*cosα - sinα) =(2cos²α -cosα) / (2cosα -1)sinα=
(2cosα -1)cosα / (2cosα -1)sinα = ctgα .
2) ( sin²α - 4sin²(α/2) ) / ( sin²α - 4+4sin²(α/2) ) =
( 4sin²(α/2) cos²(α/2) - 4sin²(α/2) ) / ( 4sin²(α/2) cos²(α/2) - 4(1 -sin²(α/2) ) =
- 4sin²(α/2) (1 - cos²(α/2) ) / - 4cos²(α/2)( 1 -sin²(α/2) ) =
sin⁴(α/2) / cos⁴(α/2)= tg⁴(α/2) .
3) (cos²α -sin²α ) / (1+sin2α) =
|| * * * 1+sin2α= cos²α+sin²α+2sinαcosα =(cosα+sinα)² * * * ||
= (cosα -sinα )(cosα+sinα) /(cosα+sinα)²= (cosα -sinα)/(cosα+sinα)
4) ( sin(α+β) - sin(α -β) ) / ( sin(α+β) +sin(α -β) ) =
|| sinα -sinβ =2sin( (α -β)/2 ) *cos( (α +β)/2 ) ||
|| sinα+sinβ =2sin( (α+β)/2 ) *cos( (α -β)/2 ) ||
= 2 sinβcosα / 2 sinαcosβ =(cosα / sinα) *(sinβ/cosβ) = ctgα *tgβ =
ctgα / ctg β.
* * * * * * * по другому * * * * * * *
( sin(α+β) - sin(α -β) ) / ( sin(α+β) +sin(α -β) ) =
( sinαcosβ+cosα*sinβ - (sinαcosβ-cosα*sinβ) ) *
1/ ( sinαcosβ+cosα*sinβ+ sinαcosβ-cosα*sinβ ) =
2cosα*sinβ /2 sinαcosβ =ctgα /ctgβ
* * * * * * * * * * * * * * * * * * * * *
5) ( (1+cosx)/sinx )*(1+ ( (1 -cosx)/sinx )² ) =
( (1+cosx)/sinx )*(sin²x +1 -2cosx+cos²x )/sin²x ) =
( (1+cosx)/sinx )*( 2(1 -cosx))/sin²x ) = 2(1+cosx)(1-cosx) /sin³x =
2(1 - cos²x) /sin³x =2sin²x/ sin³x = 2 / sinx .
* * * * * * * по другому * * * * * * *
= ( 2cos²(x/2) / 2sin(x/2)*cos(x/2) )*(1+ ( 2sin²(x/2) / 2sin(x/2)*cos(x/2) )² ) =
(cos(x/2) / sin(x/2) )*( 1 + sin²(x/2) / cos²(x/2) ) =
(cos(x/2) /sin(x/2) )*( ( cos²(x/2) + sin²(x/2) ) /cos²(x/2) ) =
(cos(x/2) /sin(x/2) )* ( 1 / cos²(x/2) ) = 1 /( cos(x/2)*sin(x/2) ) =2/sinx
* * * * * * * * * * * * * * * * * * * * *
172.
1) 5^(x+y)=125, (1)
3^((x-y)²-1)=1; (2)
5^(x+y)=5³, (1)
3^((x-y)²-1)=3^0; (2)
x+y=3, (1)
(x-y-1)(x-y+1)=0; (2)
y=3-x, (1)
(x-3+x-1)(x-3+x+1)=0; (2)
(2x-4)(2x-2)=0;
2x-4=0;
2x=4;
x1=2
или
2x-2=0;
2x=2;
x2=1.
y1=3-2=1;
y2=3-1=2.
ответ: (2;1), (1;2).
2) 3^x+3^y=12, (1)
6^(x+y)=216; (2)
6^(x+y)=6³;
x+y=3;
y=3-x;
3^x+3^(3-x)=12; (1)
3^(2x)-12*3^x+27=0;
3^x=t;
t²-12t+27=0;
D=144-108=36;
t1=(12-6)/2=3;
t2=(12+6)/2=9;
3^x=3;
x1=1;
3^x=9;
x2=2;
y1=3-1=2;
y2=3-2=1.
ответ: (1;2), (2;1).
3) 4^(x+y)=128, (1)
5^(3x-2y-3)=1; (2)
2^(2(x+y))=2^7, (1)
5^(3x-2y-3)=5^0; (2)
2x+2y=7, (1)
3x-2y-3=0; (2)
2y=7-2x, (1)
3x-7+2x-3=0; (2)
6x=10;
x=10/6=5/3;
y=(7-2x)/2=(7-10/3)/2=11/6.
ответ: (5/3;11/6).
4) 3^(2x-y)=1/81, (1)
3^(x-y+2)=27; (2)
3^(2x-y)=3^(-4), (1)
3^(x-y+2)=3³; (2)
2x-y=-3, (1)
x-y+2=3; (2)
x-y=1;
y=x-1;
2x-x+1=-3; (1)
x=-4;
y=-4-1=-5.
ответ: (-4;-5).
173.
1) 4^(x+y)=16, (1)
4^(x+2y-1)=1; (2)
4^(x+y)=4², (1)
4^(x+2y-1)=4^0; (2)
x+y=2, (1)
x+2y-1=0; (2)
y=2-x; (1)
x+2(2-x)-1=0; (2)
x+4-2x-1=0;
-x=-3;
x=3;
y=2-3=-1.
ответ: (3;-1).
2) 6^(2x-y)=√6, (1)
2^(y-2x)=1/√2; (2)
6^(2x-y)=6^(1/2); (1)
2^(y-2x)=2^(-1/2); (2)
2x-y=1/2, (1)
+
y-2x=-1/2; (2)
0=0
ответ: нет решений.
3) 5^(2x+y)=125, (1)
7^(3x-2y)=7; (2)
5^(2x+y)=5³, (1)
7^(3x-2y)=7^1; (2)
2x+y=3, (1)
3x-2y=1; (2)
y=3-2x; (1)
3x-2(3-2x)=1;
3x-6+4x=1;
7x=7;
x=1;
y=3-2*1=1.
ответ: (1;1).
4) 3^(4x-3y)=27√3, (1)
2^(4y+x)=1/(2√2); (2)
3^(4x-3y)=3^(7/2), (1)
2^(4y+x)= 2^(-3/2); (2)
4x-3y=7/2, (1)
4y+x=-3/2; (2)
x=-3/2-4y,
4(-3/2-4y)-3y=7/2; (1)
-6-16y-3y=7/2;
-19y=19/2;
y=-1/2;
x=-3/2-4(-1/2)=-3/2+2=1/2.
ответ: (1/2;-1/2).
Tg3a=-1
3a=-pi/4+pi*k
a=-pi/12+pi*k/3