We know that vector a = (cosx, SiNx), vector b = (cosy, siny), | A-B | = 2 * & 5 / 5, 1. Find the value of COS (X-Y) 2. If 0 Please write the specific steps (O & # 180; &;'O) ぉ (O & # 180; Д'o) は (O & # 180; ω'o) ょ

We know that vector a = (cosx, SiNx), vector b = (cosy, siny), | A-B | = 2 * & 5 / 5, 1. Find the value of COS (X-Y) 2. If 0 Please write the specific steps (O & # 180; &;'O) ぉ (O & # 180; Д'o) は (O & # 180; ω'o) ょ

Vector a = (cosx, SiNx), vector b = (cosy, siny),
Then a ^ 2 = 1, B ^ 2 = 1,
ab= cosx cosy+sinx siny=cos(x-y),
|a-b|=2√5/5,
The square is: | A-B | ^ 2 = 4 / 5,
That is, a ^ 2 + B ^ 2-2ab = 4 / 5,
1+1-2 cos(x-y) =4/5,
cos(x-y) =3/5.
π / 2 ＜ y ＜ 0, siny = - 5 / 13,
Then cosy = 12 / 13
0

Let a = (cosx, SiNx) B = (cosy, siny), where 0

|2A + B | = | a-2b | both sides square at the same time to get 4A ^ 2 + 4AB + B ^ 2 = a ^ 2-4ab + 4B ^ 2 (*) and a = (cosx, SiNx) B = (cosy, siny) so | a | = 1 | B | = 1, so (*) formula is simplified to 8ab = 0, that is, cosx * cosy + SiNx * siny = cos (X-Y) = 0
Another 0

Given vector a = (cosx, SiNx), vector b = (cosy, siny), if X-Y = π / 3, then the angle between vector a and vector a + B

cosθ=a*(a+b)/|a|*|a+b|=(a^2+a*b)/|a+b|
=(1+a*b)/|a+b|
a*b=cosxcosy+sinxsiny
=cos(y-x)=cos(π/3)=1/2
|a+b|^2
=|a|^2+|c|^2+2a*b
=1+1+2*1/2
=3
∵cosθ=(1+1/2)/√3=√3/2
∴θ=π/6

The vector a = (cosx, SiNx), B = (- siny, cosy), and X belongs to [π / 4, π / 2]
1: If y = 3x, find a + B
2: If y = 3x, and the inequality a * b-radical 2 * a + b * k is greater than or equal to - 3 / 2, find the value range of real number K

1) Vector a + B = (cosx siny, SiNx + cosy) = (cosx-sin3x, SiNx + cos3x)
|a+b|²=(cosx-sin3x)²+(sinx+cos3x)²
=cos²x-2cosxsin3x+sin²3x+sin²x+2sinxcos3x+cos²3x
=2-2(cosxsin3x-sinxcos3x)
=2-2sin2x
=2(sin²x+cos²x-2sinxcosx)
=2(sinx-cosx)²
Because x ∈ [π / 4, π / 2]
So SiNx > cosx
Then | a + B | = √ 2 (SiNx cosx)
2) Vector AB = - cosxsin3x + sinxcos3x = - sin2x
ab-√2|a+b|k≥-3/2
That is - sin2x-2k (SiNx cosx) ≥ - 3 / 2
-sin2x-2k√(1-sin2x)≥-3/2
1-sin2x-2k√(1-sin2x)+1/2≥0
Let √ (1-sin2x) = t, then 0 ≤ t ≤ 1
t²-2kt+1/2≥0
That is, if K ≤ (T & # 178; + 1 / 2) / 2T is constant, then K should be less than the minimum value of (T & # 178; + 1 / 2) / 2T
And (T & # 178; + 1 / 2) / 2T = 1 / 2 (T + 1 / 2t) ≥ √ 2 / 2
∴k≤√2/2

Given the function f (x) = x ^ 2-2ax, the image of function g (x) is obtained by moving the image of function f (x) one unit to the left, and y = g (x) is an even function
【1】 Finding the value of a
【2】 Let f (x) = f (x) * [g (x) + 1], find the maximum and minimum value of F (x) in the interval [1,3]!

If the image of F (x) = (x-a) ^ 2-A ^ 2 is shifted one unit to the left, the resulting function is g (x) = (x + 1-A) ^ 2-A ^ 2, because g (x) is an even function, so a = 1
F (x) = x ^ 4-2x ^ 3, and its derivative is f '= 4x ^ 2 (x-1.5). It can be seen that the original function monotonically decreases at X1.5. If we investigate the function value of x = 0, x = 1, x = 1.5, x = 3, we can get the maximum value of F (3) = 27,
The minimum value is f (1.5) = - 27 / 16

It is known that the image of function y (x) is symmetric to the image of F (x) about y = x after it is shifted one unit to the left along the x-axis,
And G (19) = a + 2, then the range of function y = 3 ^ (AX) (0 < x ≤ 1) is?
f(x)=3^x

The range is (1,2]

The image of the function y = 2x + 3 is translated so that it passes through points (2, - 1)

Let the analytic expression after translation be y = 2x + B, and substitute the point (2, - 1) to get - 1 = 4 + B ∥ B = - 5 ∥ then the analytic expression is y = 2x-5

Once the function y = - X-1, how to translate the function image to pass through the point P (2,4)?
Processes~

For a function y = - X-1, we add a letter A to X
y=-(x-a)-1,
Let this function pass P (2,4)
So, 4 = - (2-A) - 1, a = 7
If a is a negative number, then move to the right. In this topic, the line moves to the left by 7 units. We can get the result we need
Let y-b = - X-1,
The coordinates of P are also substituted into the equation
4-b=-2-1,
b=7.
That is to say, if we move the original straight line up 7 units, we can get the result we need

How to prove that the intersection of the image of the inverse scale function and the image of the positive scale function is symmetrical about the origin?

Curve symmetry should be simplified to the point-to-point problem. It can be proved by grasping this point. The main difficulty is the idea

It is known that the image of positive scale function y = KX intersects with the image of inverse scale function y = - 1 / X at point m, N. if the length of segment Mn is 2 √ 2, then K=__
To explain

Let the point m (x1, Y1) n (X2, Y2) y = kxy = - 1 / X combine to get: x ^ 2 = - 1 / K, where k is obviously less than 0. For the convenience of calculation, let - k = t (T > 0) get x ^ 2 = 1 / T, so X1 = √ T / T, X2 = - √ T / T, so Y1 = √ T, y2 = - √ TMN = √ (x1-x2) ^ 2 + (y1-y2) ^ 2) = √ (2 √ T / T) ^ 2 + (2 √ T) ^ 2 = √ (