F (x) = sin (π X / 4 - π / 6) - 2cos (π X / 8) ^ 2 + 1 (1) find the minimum positive period of F (x). ② if the image of function y = g (x) is about a pair of lines X = 1 If the graph of the function y = g (x) is related to the line x = 1 Why g (x) = f (2-x) means that G (x) and f (x) are symmetric about a point (2-x, y)

F (x) = sin (π X / 4 - π / 6) - 2cos (π X / 8) ^ 2 + 1 (1) find the minimum positive period of F (x). ② if the image of function y = g (x) is about a pair of lines X = 1 If the graph of the function y = g (x) is related to the line x = 1 Why g (x) = f (2-x) means that G (x) and f (x) are symmetric about a point (2-x, y)

Well, you can draw a diagram to show that if the point (x, y) is a point on G (x), that is, y = g (x), then the symmetric point of point (x, y) with respect to x = 1 is (2-x, y). According to the question g (x) and f (x) about x = 1, it should be on the function f (x), so y = f (2-x) can be obtained by substituting y = g (x)

Let f (x) = sin (π X / 4 - π / 6) - 2cos ^ 2 π X / 8 + 1. Find the minimum positive period of F (x)

f（x）= sin(πx/4-π/6) - [2cos^2πx/8-1] = sin(πx/4-π/6) - cosπx/4
The minimum positive period should be 2 π / (π / 4) = 8

The known function f (x) = 4cosxsin (x + π) 6)−1 (I) find the minimum positive period of F (x); (II) find f (x) in the interval [- π 6，π 4] Maximum and minimum values on

(I) ∵ f (x) = 4cosxsin (x + π 6) - 1 = 4cosx (32sinx + 12cosx) - 1 = 3sin2x + cos2x = 2Sin (2x + π 6), ∵ the minimum positive period of F (x) t = 2 π 2 = π; (II) ∵ x ∈ [- π 6, π 4], ? 2x + π 6 ∈ [- π 6, 2 π 3],  - 12 ≤ sin (2x + π 6) ≤ 1, - 1 ≤ 2Si

Given the function f (x) = 4cosxsin (x + π / 6) - 1, find the maximum and minimum values of F (x) on the interval [- π / 6, π / 4]

f(x)=4cosxsin(x+π/6)-1
=4cosx(√3/2sinx+1/2cosx)-1
=2√3sinx+2cos^2x-1
=√3sin2x+cos2x
= 2sin(2x+π/6)
If x ∈ [- π / 6, π / 4], then (2x + π / 6) ∈ [- π / 6,2 π / 3]
Draw a unit circle and it will come out in a stroke
So the maximum value of F (x) is 2 and the minimum value is - 1

Given the function f (x) = 4cosxsin (x + Pai / 6) - 1, find the minimum positive period of F and find its maximum and minimum value on the interval [- Pai / 6, Pai / 4]

From the formula of sum and difference of product, we can get the following results
f(x)=2[sin(2x+π/6)+sin(π/6)]-1=2sin(2x+π/6)
Therefore, the minimum positive period of F is t = π
Because it is obvious that the minimum value of π / 6 is f (- π / 6) = - 1

It is known that the minimum positive period of the function f (x) = 4coswx · sin (Wx + Pai / 4) (W > 0) is Pai  

(1) The minimum positive period of ∵ function f (x) = 4coswx · sin (Wx + Pai / 4) (W > 0) is Pai
f(x)=4coswx•sin(wx+pai/4)=2√2coswx•sinwx+2√2cos^2wx
=√2sin2wx+√2cos2wx+√2
=2sin(2wx+π/4)+√2
2w=2π/π==>w=1
∴f(x)=2sin(2x+π/4)+√2
(2) Analysis: ∵ f (x) = 2Sin (2x + π / 4) + √ 2
Monotone increasing region: 2K π - π / 2

Let f (x) = 2Sin (Wx - π / 6) · sin (Wx + π / 3) (where w > 0, the minimum positive period of X ∈ R is π) Let f (x) = 2Sin (Wx - π / 6) · sin (Wx + π / 3) (where w > 0, the minimum positive period of X ∈ R is π). Question: (1) find the value of W. (2) in triangle ABC, if a ＜ B and F (a) = f (b) = 1 / 2, find BC / ab

(1) f(x)=2sin(wx-π/6)•sin(wx+π/2-π/6)
=2sin[π/2+(wx-π/6)]•sin(wx-π/6)
=2cos(wx-π/6)•sin(wx-π/6)
=sin(2wx-π/3)
Because the period T = 2 π / 2W = π, then w = 1
So f (x) = sin (2x - π / 3)
In the triangle ABC, if a ＜ B, and f (a) = f (b) = 1 / 2
Then f (x) = sin (2x - π / 3) = 1 / 2,
We know that 2A - π / 3 = π / 6, a = π / 4
2x-π/3=π-π/6,B=7π/12
So C = π - A-B = π - π / 4-7 π / 12 = π / 6
From the sine theorem BC / AB = Sina / sinc
=sin(π/4)/sin(π/6)
=(√2/2)/(1/2)
=√2

The minimum positive period of the function f (x) = 4cos (w π) sin (Wx + π / 4) (W > 0) is known to be π (1). Find the value of W (2) discuss the value of F (x) in the interval [0, π / 2] Monotonicity over

The minimum positive period of the function f (x) = 4cos (w π) sin (Wx + π / 4) (W > 0) is known to be π (1). Find the value of W (2) discuss the monotonicity of F (x) on the interval [0, π / 2]
(1) Analysis: the minimum positive period of ∵ function f (x) = 4cos (w π) sin (Wx + π / 4) (W > 0) is π
∴T=π==>w=2π/π=2
f(x)=4cos(2π)sin(2x+π/4)=4sin(2x+π/4)
(2) Analysis: 2K π - π / 2 < = 2x + π / 4 < = 2K π + π / 2 = = > k π - 3 π / 8 < = x < = k π + π / 8, f (x) increases monotonically;
2K π + π / 2 < = 2x + π / 4 < = 2K π + 3 π / 2 = = > k π + π / 8 < = x < = k π + 5 π / 8, f (x) decreases monotonically;
∵ interval [0, π / 2]
ν monotonically increases on [0, π / 8] and decreases monotonically on [π / 8, π / 2];

Let f (x) = sin (Wx + φ) + cos (Wx + φ) (W > 0, | φ)|

f(x)=sin(wx+φ)+cos(wx+φ)
=√2sin(wx+φ+π/4)
T=2π/w=π
W=2
f(x)=√2sin(2x+φ+π/4)
f(-x)=f(x),
So f (- π / 8) = f (π / 8)
sinφ=sin(π/2+φ)=cosφ
tanφ=1
|φ|

The function y = 1-2cosx, [0,2 π] is drawn by "five point method", and the maximum and minimum values of this function are written out according to the image You don't need to draw an image. Just ask for the maximum and minimum values. I've calculated. Only the maximum value is 3, right?

The pink line is y = cosx. Fold it along the X axis to get y = - cosx (purple line). Double the ordinate of y = - cosx to get y = - 2cosx (red line) and y = - 2cosx