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The known function y = asin (ω x + φ), |φ|
If the period is π, then the maximum value of ω = 2 is 3, so a = 3
π / 12 × 2 + φ = π / 2 + 2K π K ∈ N and | φ|
It is known that the function y = asin (Wx + b) (a > 0, w > 0) in the same period, x = π / 9 is, and the maximum value is 1 / 2; when x = 4 π / 9, the minimum value is - 1 / 2,
Absolute value of the best value = 1 / 2
A 〉 0, so a = 1 / 2
The next pair of maxima and minima is half a cycle
T/2=4π/9-π/9=π/3
T=2π/3
|w|=2π/T
W>0
So w = 3
y=1/2sin(3x+b)
x=π/9,y=1/2
sin(π/3+b)=1=sin(π/2)
b=π/6
y=1/2sin(3x+π/6)
It is known that the function f (x) = asin (Wx + φ) + n has a maximum value of 4, a minimum value of 0 and a minimum positive period of Π,
∵ the maximum value is 4, the minimum value is 0, ᙽ the amplitude is a = (4-0) ∵ 2 = 2 ∵ the maximum value is 4 = n + A,
The function y = asin (Wx + φ) + B (a > 0, w > 0, absolute value φ ≤ π) when x = π / 6, y takes the minimum value of 1; when x = 5 π / 6, y takes the maximum value of 3, and calculates the function analysis
When x = π / 6, y gets the minimum value of 1, so sin (Wx + φ) = - 1, so we get - A + B = 1
When x = π 5 / 6, y reaches the maximum value of 3, so sin (Wx + φ) = 1, and then a + B = 3
A = 1, B = 2
Note that the solution of W is not unique
Here we find that when x = π / 6 to 5 π / 6, it happens to be a period
The function y = asin (Wx + φ) + B (a > 0, w > 0, absolute value φ ≤ π) when x = π / 6, y takes the minimum value of 1; when x = 5 π / 6, y takes the maximum value of 5, and calculates the function analysis The function y = asin (Wx + φ) + B (a > 0, w > 0, absolute value φ ≤ π) is in a period
The solution is: a = 2, B = 3 when x = π / 6, when x = π / 6, when x = π / 6, when x = π / 6, when x = π / 6, y takes the minimum value 1; when x = 5 π / 6, y takes the maximum value; when x = 5 π / 6, y takes the maximum value; when x = 5 π / 6, y takes the maximum value; when x = 5 π / 6, y takes the maximum value, 5 ɀ periodic T / 2 π / 6-π / 6 = 2 π / 3T = 4 π / 3 = 3 = 2: B-A = 5: B + a = 5: a = 2, B = 3: when x = π / 6, when x = π / 6, when x = 5 π / = k π - π /
Given the function y = asin (Wx + φ) + B (a > 0, w > 0, absolute value φ ≤ π), when x = π / 6, y takes the minimum value 1, and the minimum positive period of this function is 4 π / 3 and the maximum value
T=4π/3=2π/ω ω=3/2
When x = π / 6, the minimum value of Y is 1
-A + B = 1, and sin [(3 / 2) (π / 6) + φ] = - 1
That is, π / 4 + φ = 2K π - π / 2
φ = 2K π - 3 π / 4, k = 0, φ = - 3 π / 4
The conditions are not enough
If there is a maximum value of M
Then there are
A+b=m
A=（m-1)/2,b=(m+1)/2
The problem is solvable
The function y = asin (Wx + b) (a > 0, w > 0) takes a maximum value and a minimum value in X ∈ (0,7 π), and when x = π, y has a maximum value of 3, and when x = 6 π, y has a minimum value - 3 (1). Find the analytic formula of this function (2) write the monotone increasing interval of the function
According to the graph of trigonometric function, we know that the difference between the maximum value and the minimum value is a period, so 6 π - π = 5 π = t / 2
So t = 10 π,
W = 2 π / T, so w = 1 / 5
The maximum and minimum values show that a = 3
By introducing x = π into the equation, 3 = 3sin (0.2 π + b) is obtained
So B + 0.2 π = 0.5 π, so B = 0.3 π
So the explanation is y = 3sin (0.2x + 0.3 π)
Then the monotone increasing intervals are (0, π) and (6 π, 7 π)
The function y = asin (ω x + φ) (a > 0, ω > 0) in the same period, when x = π / 3, the maximum value of y = 2, and when x = 0, the minimum value of y = - 2 When x = π / 3, y max = 2, then a = 2, ω = 3 How is Omega calculated?
When x = O, φ = - π / 2
ω x + φ = π / 2
The maximum value of the function y = asin (ω x + φ) + n is known to be 4, the minimum value is 0, the minimum positive period is π / 2, and the straight line x = π / 3 If a ≠ 0, ω > 0, - π / 2
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
y=2sin(4x+π/6）+2
The known function f (x) = asin (Wx + a), where w > 0, | a | π / 2 (1) If cos π / 4 cosa-sin3 π / 4 Sina = 0, find the value of A (2) Under the condition of (1), if the distance between two adjacent symmetry axes of the image of function f (x) is equal to π/ 3. Find the analytic formula of function f (x), and find the minimum positive real number m, so that the image of function f (x) is shifted to the left by M units, and the corresponding function is even function
1. SIN3 π / 4 = sin π / 4cos π / 4cosa sin π / 4sina = cos (π / 4 + a) = 0A = π / 42, the period is 2 π / 3W = 3, so the analytic formula is f (x) = asin (3x + π / 4) left shift m, that is, asin (3 (x + m) + π / 4) = asin (3m + π / 4 + 3x) is even function, then 3M + π / 4 is even function
RELATED INFORMATIONS
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- 3. How to learn the properties of the function y = asin (Wx + φ)? It feels so difficult
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