It is known that the function y = asin (ω x + φ) has a maximum value of 2 when x = π - 3 and a minimum value of - 2 when x = 0 in the same period

It is known that the function y = asin (ω x + φ) has a maximum value of 2 when x = π - 3 and a minimum value of - 2 when x = 0 in the same period

A=2 π∕3x＋α＝π／2 α＝2π／3

It is known that the function y = asin (Wx + b) has a maximum value of 2 when x = 3 / π and a minimum value of - 2 when x = 0,

y=Asin(ωx+b)
-1≤sin(ωx+b)≤1
1. When a > 0:
① Only when sin (ω x + b) = 1, y gets the maximum value a
The maximum value of Y is known to be 2
Therefore, a = 2
In this case, sin (ω x + b) = 1
In the minimum positive period, there is: ω x + B = π / 2
It is known that x = π / 3
That is: ω π / 3 + B = π / 2 (1)
② Only when sin (ω x + b) = - 1, y gets the minimum value - a = - 2
In this case, sin (ω x + b) = - 1
In the minimum positive period, there is: ω x + B = 3 π / 2
Known: x = 0
That is: ω × 0 + B = 3 π / 2
The solution is: B = 3 π / 2
Substituting (1), there is: ω π / 3 + 3 π / 2 = π / 2
The solution is: ω = - 3
The analytical formula is: y = 2Sin (- 3x + 3 π / 2)
2. When a < 0:
① Only when sin (ω x + b) = - 1, y gets the maximum value a
The maximum value of Y is known to be 2
Therefore, a = - 2
In this case, sin (ω x + b) = - 1
In the minimum positive period, there is: ω x + B = 3 π / 2
It is known that x = π / 3
That is: ω π / 3 + B = 3 π / 2 (2)
② Only when sin (ω x + b) = 1, y gets the minimum value a = - 2
In this case, sin (ω x + b) = 1
In the minimum positive period, there is: ω x + B = π / 2
Known: x = 0
That is: ω × 0 + B = π / 2
The solution is: B = π / 2
Substituting (2), there is: ω π / 3 + π / 2 = 3 π / 2
The solution is: ω = 3
The analytic formula is y = - 2Sin (3x + π / 2)
To sum up, there are two simplest expressions of the analytic formula
Y = 2Sin (- 3x + 3 π / 2), and: y = - 2Sin (3x + π / 2)
One more sentence:
In fact, the above two simplest analytic expressions are equivalent

The known function y = asin (Wx + φ), |φ|

When x = π / 12, the maximum value is 3, and when x = 7 π / 12, the minimum value is - 3
We get a = 3 T / 2 = 7 π / 12 - π / 12, so t = π w = 2
π/12*2+φ=kπ+π/2 ,|φ|

In the same period, the function y = asin (ω x + φ) obtains the maximum value of 1 / 2 when x = π / 9, and the minimum value - 1 / 2 when x = 4 π / 9 in the same period The answer is 1 / 2 sin (3x + π / 6) I want to ask why there is only one answer |When ω| = 3 takes - 3, can we calculate the solution of φ = 5 π / 6? There is no such regulation

In fact, the equation: y = asin (ω x + φ) is derived from the simple harmonic vibration in physics. For a simple harmonic vibration, the number of times the object completes the full vibration in unit time is called frequency, expressed by F, 2 π times of frequency is called angular frequency, that is, ω = 2 π F, so by default, ω > 0

It is known that the maximum value of the function f (x) = asin ^ 2 (Wx + ψ) is 2, and the distance between the two symmetry axes of the image is 2 It's urgent tonight

f(x)=Asin^2(wx+ψ)=A*[1-cos2(wx+ψ)]/2
Maximum value 2: a = 2
Then the period is 4,2 π / 2W = 4, w = π / 4
2 * [1-cos2 (π / 4 + ψ)] / 2 = 2
Cos2 (π / 4 + ψ) = - 1, that is cos (π / 2 + 2 ψ) = - sin2 ψ = - 1
If sin 2 ψ = 1, then 2 ψ = k π + π / 2, ψ = k π / 2 + π / 4

Function y = asin (ω x + φ) (a > 0, ω > 0, 0 ≤ φ ≤ π 2) In X ∈ (0,7 π), only one maximum value and one minimum value are obtained, and when x = π, ymax = 3; when x = 6 π, Ymin = - 3 (1) Find the analytic expression of this function; (2) Find the monotone interval

(1) From the meaning of the title, a = 3, period T = 2 (6 π - π) = 10 π = 2 π
ω，∴ω=1
5．
According to the point (π, 3) on the graph of the function, 3sin (1) can be obtained
5 π + φ) = 3, sin (π) can be obtained
5+φ）=1．
Binding 0 ≤ φ≤ π
2, we can get that φ = 3 π
The analytic expression of the function is y = 3sin (1
5x+3π
10）．
(2) Let 2K π - π
2≤1
5x+3π
10≤2kπ+π
2, K ∈ Z, we obtain 10K π - 4 π ≤ x ≤ 10K π + π, K ∈ Z,
So the increasing interval of the function is [10K π - 4 π, 10K π + π], K ∈ Z

Function y = asin (Wx + ∮) (a, w > 0,0 (3) Whether there is a real number m, satisfying the inequality: asin (ω √ (- m ^ 2 + 2m + 3) + φ) > asin (ω √ (- m ^ 2 + 4) + φ? If it exists, find the value (or range) of M, if not, please explain the reason

(10kπ-4π,10kπ+2π)

The function y = asin (ω x + Φ) (a > 0, ω > 0) in the same period, when x = π / 12, y max = 2; when x = 5 π / 12, y min = - 2 Find expression

Because the maximum and minimum is 2, - 2
So a = 2
wπ/12 +Φ =π/2 +2kπ
w5π/12+Φ=3π/2+2kπ
w=3 Φ=π/4
y=2sin(3x+π/4)

It is known that the maximum value of the function y = asin (ω x + φ) + n is 4, the minimum value is 0, the minimum positive period is π / 2, and the straight line x = π / 3 How to find the value of φ?

The median of the maximum and minimum is 2
So n = 2
Max min = 4
So amplitude = 4 / 2 = 2
T=π/2=2π/w
W=4
y=2sin(4x+φ）+2
Axis of symmetry x = π / 3
So sin (4 π / 3 + φ) = ± 1
φ=π/6

It is known that the image of function f (x) = asin (Wx + φ) (a > 0, w > 0,0 ＜ π / 2) is symmetric with respect to point B (- π / 4,0). The shortest distance between point B and the symmetry axis of the image with function y = f (x) is π / 2, and f (π / 2) = 1 1. The first question is very simple 2. If 0 < θ < π, and f (θ) = 1 / 3, find the value of Cos2 θ Help, do it. There's another reward.

The shortest distance from the symmetric center B to the symmetry axis of the image of the function image is π / 2, so t / 4 = π / 2T = 2 π, so w = 10 = asin (- π / 4 + 4 + φ) so - π / 4 + φ = 0, φ = 0, φ = π / 4f (π / 2) = asin (π / 2 + 2 + π / 4) = 1A = √ 2, w = 1, w = 1, φ = π / 4f (θ) = √ 2Sin (θ + π / 4) = 1 / 3sin (θ + π / 4) = 1 / 3ssin (θ + π / 4) = √ 2 / 6sin θ cos cos cos cos cos cos cos cos cos (2 / 6sin) cos cos cos cos cos cos cos (2 / 6sin)