Given the function f (x) = 2sinwx-2cos ^ 2wx (x belongs to R, w > 0), the distance between two adjacent symmetry axes of the image of F (x) is equal to π / 2 1. Finding the minimum positive period of function f (x) 2. Finding the value of F (π / 4)

Given the function f (x) = 2sinwx-2cos ^ 2wx (x belongs to R, w > 0), the distance between two adjacent symmetry axes of the image of F (x) is equal to π / 2 1. Finding the minimum positive period of function f (x) 2. Finding the value of F (π / 4)

T=2*π/2=π
f(x)=sin2ωx-cos2ωx-1=2sin(2ωx-π/4)-1．
Because t / 2 = π / 2, t = π, ω = 1
So f (x) = 2Sin (2x - π / 4) - 1
So f (π / 4) = 0 (7 points)

The nearest distance between two symmetry axes of image with function y = 2cos (1 / 3x + π / 3) + 1

3∏

The distance between two adjacent symmetric axes of function f (x) = 2cos2x-1 is ()
A. 2πB. πC. π2D. π4

The function f (x) = 2cos2x-1 = cos2x, the period T of the function is 2 π 2 = π, because the distance between two adjacent symmetrical axes is half of the period, that is, π 2, then the distance between two adjacent symmetrical axes of the function is π 2

Find the center and axis of symmetry of the function y = 2cos (- 3x + Π / 3), and find the maximum, minimum and monotone interval when x ∈ [- Π / 3, Π / 2]

Y = 2cos (- 3x + Π / 3) = 2cos (3x - Π / 3). So the maximum value is 2 and the minimum value is - 2. Take 3x - Π / 3 as a whole = k Π, and remove x as the axis of symmetry. Take 3x - Π / 3 as a whole = Π / 2 + K Π, and solve x as the center of symmetry

Given that the minimum positive period of the function f (x) = 2cos ^ 2wx + 2sinwx × cosxwx + 1 (x ∈ R, w > 0) is π / 2, (1) find the value of W, (2) find the maximum value of X

2sinwx×cosxwx=sin2wx
2cos^2wx=cos2wx+1
F (x) = cos2wx + sin2wx + 2 = radical 2 * sin (2wx + π / 4) + 2
T=π/2
w=2
F (x) = radical 2 * sin (4x + π / 4) + 2
Function maximum
x=k*π/2+π/16

(1 / 2) the known function f (x) = 2cos ^ 2wx + 2 √ 3sinwxcoswx (where 0

F (x) = cos2wx + √ 3sis2wx + 1 = 2Sin (2wx + π / 6) + 1,
So, w = 1 / 2

It is known that the minimum positive period of function f (x) = 2cos2 ω x + 23sin ω xcos ω X-1 (ω > 0) is π. (1) find the value of F (π 3); (2) find the monotone increasing interval of function f (x) and the equation of symmetry axis of its image

(1) The function f (x) = 2cos2 ω x + 23sin ω xcos ω X-1 = Cos2 ω x + 3sin2 ω x = 2Sin (2 ω x + π 6), because the minimum positive period of F (x) is π, so 2 π 2 ω = π, the solution is ω = 1, so f (x) = 2Sin (2x + π 6), f (π 3) = 2sin5 π 6 = 1

If the point P (0,1) is on the graph of the function y = x ^ 2 + BX + C and f (x) = 3, then the equation of the symmetry axis of the graph of the function is a process of () writing

Substituting the point P (0,1) into the function y = x ^ 2 + BX + C, C = 1
But how much is x in F (x) = 3

Y = f (x) has f (x + a) = f (b-X), why is the axis of symmetry x = (a + b) / 2, y = f (x + a) and y = f (b-X) about x = (B-A) / 2

On the problem of the axis of symmetry: F (x) is symmetric with respect to x = a, then if (M + n) = 2A, then f (m) = f (n) this is the most basic, any problem about the axis of symmetry should start from here. Because for any x, f (x + a) = f (b-X), a and B are constants, so (x + a) + (b-X) is twice the axis of symmetry, so

If the image of function y = f (x) has two symmetry axes perpendicular to X axis, it is proved that y = f (x) is a periodic function

The image of the function y = f (x) has two symmetry axes perpendicular to the X axis, which are x = X1 and x = x2
For x = X1 symmetry, then f (x) = f (2x1-x)
For x = x2 symmetry, f (2x1-x) = f (2x2 - (2x1-x)) = f (x + 2x2-2x1)
∴f(x)=f(x+2x2-2x1)
Ψ y = f (x) is a periodic function