How to learn the properties of the function y = asin (Wx + φ)? It feels so difficult

How to learn the properties of the function y = asin (Wx + φ)? It feels so difficult

Learning function f (x) = asin (ω x + φ) first of all, to master the meaning of each quantity in the function: A: amplitude, the maximum value of the function is a, the minimum value is - A; ω: angular frequency, ω = 2 π / T, t: function period, φ: initial phase, that is, the phase when x = 0, ω x + φ: phase, that is, the position of the function on the X axis at x time

How to simplify the function to y = asin (Wx + ρ)? For example: how does the function y = SiNx + sin (x + π / 2) become y = asin (Wx + ρ)? How to find the maximum and minimum? For example, what is the maximum and minimum value of y = 3sin (2x + 3 / π)? The general method of finding the maximum and minimum value and the method of simplifying the function to the form of y = asin (Wx + ρ)!

How does the function y = SiNx + sin (x + π / 2) become the form of y = asin (Wx + ρ) y = SiNx + sin (x + π / 2)
=sinx+cosx
=√2(√2/2*sinx+√2/2cosx)
=√2(sinxcosπ/4+cosxsinπ/4)
=√2sin(x+π/4)
The maximum and minimum of SiNx are 1 and - 1 respectively, that is, the maximum and minimum of sin (x + π / 4) are 1 and - 1 respectively
So the maximum y = √ 2Sin (x + π / 4) is √ 2 × 1 = √ 2, and the minimum is √ 2 × (- 1) = - √ 2
How to find the maximum and minimum value of y = 3sin (2x + 3 / π)
The maximum and minimum of SiNx are 1 and - 1 respectively,
The maximum and minimum of sin (x + π / 4) are 1 and - 1 respectively
So the maximum value of = 3sin (2x + 3 / π) is 3 × 1 = 3
The minimum is 3 × (- 1) = - 3
Hope you can understand, hope to adopt, agree, thank you!

The derivation of period of function y = asin (Wx + y) and function y = ACOS (Wx + y) There are some doubts. You can say it first, and then I will ask after that

How can I change the equation to y = asin (Wx + &) and y = ACOS (Wx + &)
f(x)=Asin(wx+&)=Asin(wx+2π+&)=Asin[w(x+2π/w)+&]=f(x+2π/w)
So, the period is t = 2 π / W;
The other one is quite similar

The function y = asin (Wx + φ) + B has the highest point (π / 12,3) and the lowest point (7 π / 12, - 5) in the same period If you want the detailed process, thank you

Period T = (7 π / 12 - π / 12) * 2 = π
T=2π/w=π
W=2
A=(3-(-5))/2=4
b=(-5+3)/2=-1
2*π/12+φ=π/2+2kπ
φ=π/3+2kπ
Phi should have a range
For example: | φ| < π / 2
Then φ = π / 3
y=4sin(2x+π/3)-1

Given the function y = asin (Wx + T) + B, (a > 0, w > 0, 0 ≤ T)

b=(-3+1)/2=-1
A=[1-(-3)]/2=2
T=(7π／12-π／12)×2=π
W=2π/π=2
When x = π / 12, y = 1
∴2sin(2×π／12+t)-1=1
∴sin(π/6+t)=1
∴π/6+t=2kπ+π/2
∵0≤t

The period T of the function y = asin (Wx + b) or y = ACOS (Wx + b) (W > 0 and constant) is t = 2 π / W. why is w > 0?

It's for convenience

Function y = asin (Wx + φ) (a > 0, w > 0, | φ)|

A = 3
Two adjacent zeros π / 3,5 π / 6 are half a period apart
∴T/2=5π/6-π/3=π/2,∴T=π
From 2 π / w = π, w = 2
f(x)=3sin(2x+φ)
When x = (π / 3 + 5 π / 6) / 2 = 7 π / 12, the function gets the minimum value
∴2×7π/12+φ=2kπ+3π/2,k∈Z
∴φ=2kπ+π/3,k∈Z
∵|φ|

The known function y = asin (Wx + Q) x ∈ R (where a ＞ 0, w ＞ 0) The highest point on the image is (2,2 times the root 2) the intersection point (6,0) of the image from the highest point to the adjacent lowest point and the X axis is obtained

Max = a = 2 √ 2
The highest to the lowest is 1 / 2 cycle
The highest to the intersection of the x-axis is the highest to the lowest half, that is, 1 / 4 cycle
So t / 4 = 6-2
T=16
T=2π/w=16
w=π/8
Put (2,2 √ 2) in
2√2=2√2sin(π/4+q)
π/4+q=π/2
q=π/4
So y = 2 √ 2Sin (π X / 8 + π / 4)

The image crossing point P (π / 12,0) of the function f (x) = asin (Wx + φ), (a > 0, w > 0, | φ < π) is known The highest point nearest to the point P in the function image is (π / 3,5) (2) The value range of X satisfying f (x) < = 0

1、
If the highest point is (π / 3,5), then a = 5
The image passes through point P (π / 12,0), and the highest point closest to point P in the function image is (π / 3,5)
Then t / 4 = π / 3 - π / 12 = π / 4
Then: T = π = 2 π / W
Results: w = 2
Therefore, f (x) = 5sin (2x + φ)
By substituting the point P (π / 12,0), we get: 5sin (π / 6 + φ) = 0
Because|

The image crossing point P (π / 12,0) of the function f (x) = asin (Wx + φ) (a > 0, w > 0, |φ| < π / 2) is known And the lowest point nearest to point P is Q (- π / 6, - 2) (1) Find the analytic formula of function f (x)

F (x) = asin (Wx + φ) (a > 0, w > 0, |||||||||||||||||||||||||||,
∴wπ/12+φ=kπ,k∈Z,
The lowest point nearest to point P is Q (- π / 6, - 2),
∴-wπ/6+φ=(k-1/2)π,A=2,
By subtracting w π / 4 = π / 2, w = 2,
∴φ=（k-1/6)π,|φ|<π/2,
∴k=0,φ=-π/6,
∴f(x)=2sin(2x-π/6).