The function y = asin (ω x + φ) + m (a > 0, ω > 0, | φ| π) is known 2) Its maximum value is 4, its minimum value is 0, and its minimum positive period is π 2, the line x = π If 3 is a symmetric axis of the image, then the analytic expression of the function satisfying the condition is______ .

The function y = asin (ω x + φ) + m (a > 0, ω > 0, | φ| π) is known 2) Its maximum value is 4, its minimum value is 0, and its minimum positive period is π 2, the line x = π If 3 is a symmetric axis of the image, then the analytic expression of the function satisfying the condition is______ .

From the meaning of the title, we can get a + M = 4, A-M = 0, a = 2, M = 2
Then the minimum positive period is π
2, we can get 2 π
ω=π
2, the solution is ω = 4,
The function y = asin (ω x + φ) + M = 2Sin (4x + φ) + 2
Then x = π
3 is a symmetry axis of the image, and 4 × π can be obtained
3+φ=kπ+π
2, K ∈ Z, and | φ| π
2，
∴φ=π
6，
So the analytic formula of the function is y = 2Sin (4x + π)
6）+2，
So the answer is y = 2Sin (4x + π)
6）+2．

The function f (x) = asin ^ 2 (Wx + FAI) (a > 0, w > 0,0) is known

If the maximum value of y = f (x) is 2, a = 2
If the distance between two adjacent symmetry axes is 2, the period is 4, f (x) = asin ^ 2 (Wx + FAI) = a [1-cos (2wx + 2fai)] / 2
2π/2w=4 w=π/4
F (x) = 2Sin ^ 2 (π / 4x + FAI) substituting (1,2)
2=2sin^2(π/4+fai)
Sin (π / 4 + FAI) = 1 or - 1
And 0

When Pai / 2 is known, the maximum value of Pai / 9 is obtained in the same process,

Given the function y = asin (Wx + a), in the same period, when x = Pai / 9, the function gets the maximum value of 2, when x = 4pai / 9, the function gets the minimum value - 2,
T is the minimum period of a function
The abscissa difference between the maximum value and the minimum value is half a period
So t / 2 = 4pai / 9-pai / 9 = Pai / 3

It is known that the function FX = asin (x + ψ) (a > 0,0 ＜ψ ＜ π) x belongs to the maximum value of R is 1. The analytic formula of FX is obtained by the image crossing point m (π / 3,1 / 2)

The maximum is a = 1
Over M
sin(π/3+ψ)=1/2
π/3+ψ=5π/6
ψ=π/2
So f (x) = sin (x + π / 2)
That is, f (x) = cosx

It is known that the function f (x) = asin (x + φ) (a > 0,0 ＜Φ＜ π) x belongs to the maximum value of R, where the image passes through the point m (π / 3,1 / 2) 1 to find the analytic formula of F (x) 2. It is known that α, β belong to (0,90 °) and f (α) = 0.6f (β) = 12 / 13, find f (α - β)=

1. The maximum value is 1 = > A = 1. The maximum value is 1 = > A = 1, with points into the 1 / 2 = sin (π / 3 + 3 + φ) = > φ = π / 2F (x) = sin (x + π / 2) = - cosx2.f (α - β) = - cos (α - β) = - [12 / 13 * 12 / (13 * 0.6) + (1-cos ^ 2 α) ^ (1 / 2) * (1-cos ^ 2 β) ^ (1 / 2)] = - (12 ^ 2 * 5 / 13 ^ 2 * 3 + 5 / 13 * [1 - {12 / (13 * 0.6)} ^ (α - β) ^ (1-cos ^ 2 α 2 α) α (1 / 2 * [1 - {12 / (13 * 0.6) / (13 * 0 1 / 2

It is known that the minimum positive period of the function f (x) = sin ^ 2wx + √ 3sinwxsin (Wx + π / 2) (W > 0) is π 1) Find the value of W 2) Find the value of function f (x) on the interval [0,2 π / 3]

f(x)=(1-cos2wx)/2+√3/2*sin2x
=(√3/2)sin2wx-1/2*cos2wx+1/2
=√[(√3/2)^2+(1/2)^2]*sin(2wx-z)+1/2
Where Tanz = (1 / 2) / (√ 3 / 2)
So z = π / 6
So f (x) = sin (2wx - π / 6 / 6) + 1 / 2
T=2π/|2w|=π
W>0
So w = 1
f(x)=sin(2x-π/6)+1/2
Zero

It is known that the minimum positive period of the function f (x) = sin ^ 2wx + root sign 3sinwxsin (Wx + π / 2) is π. When x belongs to [- π / 12, π / 2], the value range of F (x) is given

sin(wx+π/2)=sinwxcosπ/2+coswxsinπ/2=coswx
F (x) = sin ^ 2wx + radical 3sinwxsin (Wx + π / 2)
=sin²wx+√3sinwxcoswx
=(1-cos 2wx)/2+√3/2sin 2wx
=1/2-1/2cos 2wx+√3/2sin2wx
=1/2+sin(2wx-30)
T=2π/(2W)=π
W=1
f(x)=1/2+sin(2x-30)
X belongs to [- π / 12, π / 2]
-π/3≤(2x-30)≤5π/6
-√3/2≤sin(2wx-30)≤1
(1-√3)/2≤f(x)≤3/2

Let f (x) = sin2wx + Radix 3sinwxsin (Wx + Pai divided by 2) (W > 0) have the minimum positive period of PI

W = 1, f (x) = sin2wx + √ 3sinwxsin (Wx + 2 / Pie) = (2 - √ 3 / 2) sin2wx, because the minimum positive period is Pai, w = 1

It is known that the minimum positive period of the function f (x) = (sinwx) ^ 2 + radical 3sinwxsin (Wx + π / 2) is π. W greater than 0 Find f (x) (2) Find the value range of function f (x) on the interval [0,2 / 3 π]

（1）f(x)=sin²ωx-√3sinωxsin(ωx+π/2)
=1/2-(1/2)cos2ωx-[(√3)/2]sin2ωx
=1/2-sin(2ωx+π/6)
T=2π/2ω=π
ω=1
f(x)=1/2-sin(2x+π/6)
(2) The monotone increasing interval of the function is (K π - π / 3, K π + π / 6)
The decreasing interval is (K π + π / 6, K π + 2 π / 3)
So when x = 2 π / 3, there is a maximum of 3 / 2
When x = π / 6, there is a minimum of - 1 / 2

It is known that the maximum value of the function f (x) = ACOS ^ 2 ω x + sin ω x · cos ω X-1 / 2 (W > 0. A > 0) is two-thirds of the root sign, and its minimum positive period is π Xie write the equation of symmetry axis of curve f (x) and the coordinates of its symmetry center

If f (x) = 1 / 2acos2wx + 1 / 2sin2wx = √ (a? + 1) / 2 sin (2wx + θ), where sin θ = A / √ (a? + 1), so √ 2 / 2 = √ (a? + 1) / 2, a = 12wx + θ = 2K π + π / 2, we get x = [2K π + π / 2-arcsin (A / √ (a? + 1))] / 2W is called axis equation center coordinate (= [2K