Given the vector a = (2sinwx, coswx + sinwx), B (comwx, coswx sinwx) (x > 0), the function f (x) = a * B, and the nearest small positive cycle of function f (x) Given the vector a = (2sinwx, coswx + sinwx), B (comwx, coswx sinwx) (x > 0), the function f (x) = a * B, and the nearest small positive period of function f (x) is Pai. Find the analytic expression of function f (x); find the monotone interval of function f (x) on [0, Pai / 2]

Given the vector a = (2sinwx, coswx + sinwx), B (comwx, coswx sinwx) (x > 0), the function f (x) = a * B, and the nearest small positive cycle of function f (x) Given the vector a = (2sinwx, coswx + sinwx), B (comwx, coswx sinwx) (x > 0), the function f (x) = a * B, and the nearest small positive period of function f (x) is Pai. Find the analytic expression of function f (x); find the monotone interval of function f (x) on [0, Pai / 2]

f(x)=a*b
=(2sinwx,coswx+sinwx)*(coswx,coswx-sinwx)
=(2sinwx)*(coswx)+(coswx+sinwx)*(coswx-sinwx)
=2sinwxcoswx+cos²wx-sin²wx
=sin2wx+cos2wx
=√2sin[2wx+(π/4)]
f(x)=√2sin[2x+(π/4)]
Find intersection, union known a = {x │ y = √ (1-2x) + (2x-1) / √ (x + 2)}, B = {y │ y = x ^ 2-2x-1} find a ∩ B and a ∪ B
It is known that a = {x │ y = 1-2x under the quadratic root sign) + 2x-1 under the quadratic root sign, where the function of Y has two terms, the former is
Under the quadratic root sign (1-2x), the last term is fraction, the numerator is (2x-1), the denominator is under the quadratic root sign (x + 2), B = {y │ y = x ^ 2
The trial interval is a ∩ B and a ∪ B
The a set here is a little unclear. The element seems to be x, and the following expression contains x to express y
The element X in a, only the range of X
Then 1-2x > = 0 and X + 2 > 0
Get - 2
Given the vector a = (sinwx, sinwx), B = (sinwx, - coswx), (W > 0), the minimum positive period of the function f (x) = a * B is π / 2?
I simplify it to f (x) = 1 / 2 - √ 2 / 2Sin (2wx + π / 4) and then t = 2 π / 2W to make it equal to π / 2. Then, do we need to add an absolute value to 2W, and then discuss it by classification, and how to do it?
If w > 0, t = 2 π / | 2W | = π / w = π / 2, w = 2
So f (x) = 1 / 2 - √ 2 / 2Sin (4x + π / 4)
When 4x + π / 4 = 2K π + π / 2, sin (4x + π / 4) = 1, f (x) has a minimum value of 1 / 2 - √ 2 / 2,
In this case, the set of X is {x | x = k π / 2 + π / 16, K ∈ Z},
When 4x + π / 4 = 2K π - π / 2, sin (4x + π / 4) = - 1, f (x) has a maximum value of 1 / 2 + √ 2 / 2,
In this case, the set of X is {x | x = k π / 2 - 3 π / 16, K ∈ Z}
Calculation: (- 2x ^ 2Y) ^ 3 + (3x ^ 2) ^ 2 * (- x ^ 2) * y ^ 3=
（-2x^2y）^3+(3x^2)^2*(-x^2)*y^3=-8x^6y^3+9x^4*(-x^2)y^3=-17x^6y^3
It is known that the minimum positive period of vector a = (sinwx, coswx), B = (coswx, √ 3coswx) (ω ＞ 0) and function f (x) = a × B - √ 3 / 2 is Wu
Find the monotone increasing interval of function f (x)
If the angles of three sides a, B and C of triangle ABC are a, B and C respectively, and satisfy B ^ 2 + C ^ 2 = a ^ 2 + √ 3bC, find the value of F (a)
1.ab=sinwxcoswx+√3cos²wx=1/2sin2wx+√3/2cos2wx+√3/2
f(x)=ab-√3/2=1/2sin2wx+√3/2cos2wx=sin(2wx+π/3)
T = π, so 2 π / 2W = π, so w = 1, so f (x) = sin (2x + π / 3),
So the monotone increasing interval is [K π - 5 π / 12, K π + π / 12]
2.cosA=(b^2+c^2-a^2)/2bc=√3/2 A=π/6
f(A)=f(π/6)=sin(2*π/6+π/3)=sin(2π/3)=√3/2
The following calculation results are correct ()
A. 3x2-2x2=1B. 3x2+2x2=5x4C. 3x2y-3yx2=0D. 4x+y=4xy
A. 3x2-2x2 = X2, so this option is wrong; B, 3x2 + 2x2 = 5x2, so this option is wrong; C, 3x2y-3yx2 = 3x2y-3x2y = 0, so this option is correct; D, 4x and y are not of the same kind, and cannot be merged. So this option is wrong; so select C
The image of the function y = sin (3x - π / 4) is symmetrical about the point (π / 3,0) after translation according to a = (m, 0). When | m | is the smallest, the vector a =?
The period of y = sin [3 (x - π / 12)] is 2 π / 3,
About (π / 12,0) symmetry, also about (π / 12 + π / 3,0) symmetry (i.e. about (5 π / 12,0) symmetry),
Every half period is a point of symmetry
If we want to minimize the absolute value of M, then M = - π / 12
After the image is translated according to a = (- π / 12,0), the symmetrical points are all translated to the left by π / 12, and the original symmetrical points (5 π / 12,0) are translated to (π / 3,0)
3x ＾ 2Y ^ 2 - [5xy ^ 2 - (4xy ＾ 2-3) + 2x ^ 2Y ^ 2, where x = - 3, y = 2
3x^2y^2--[5xy^2--(4xy^2--3)+2x^2y^2]
=3x^2y^2--[5xy^2--4xy^2+3+2x^2y^2]
=3x^2y^2--5xy^2+4xy^2--3--2x^2y^2
=x^2y^2--xy^2--3
=xy^2(x--1)--3
When x = -- 3, y = 2,
The value of the original formula = (- 3) x2 ^ 2x (- - 3 -- 1) -- 3
=(--3)X4X(--4)--3
=48--3
=45.
The analytic expression of the image translated by the function y = LG (3x-2) + 1 is y = lg3x, and the vector a is obtained
Let P (x, y), on y = LG (3x-2) + 1, point vector a (h, K). Point P '(x', y ') on y = lg3x, then x' = x + H, y '= y + K.X = x' - H, y = y '- K.Y' - k = LG [3 (X-H) - 2] + 1. Y '- k-1 = LG (3x-3h-2). And y = lg3x, then - k-1 = 0, - 3h-2 = 0, k = - 1, H = - 2 / 3. Vector a = (- 2 / 3, - 1)
2/3x^4y^8z^3-(-xy)^3(x^2y^4z^3)^2/(-1/2xyz)^3
Original formula = 2 / 3x ^ 4Y ^ 8Z ^ 3 + X & # 179; Y & # 179; (x ^ 4 y ^ 8 Z ^ 6) / [(- 1 / 8) x & # 179; Y & # 179; Z & # 179;] = 2 / 3x ^ 4Y ^ 8Z ^ 3 - 8x & # 179; Y & # 179; (XY ^ 5 Z ^ 3) = 2 / 3x ^ 4Y ^ 8Z ^ 3 - 8x ^ 4 y ^ 8 Z ^ 3 = (- 22 / 3) x ^ 4Y ^ 8Z ^ 3