Periodic maximum monotone interval y = sin (2x + π / 3) y = 2Sin (2x + π / 6) - 2 y = sin (π / 3-2x) y = Tan (π / 3-2x)

Periodic maximum monotone interval y = sin (2x + π / 3) y = 2Sin (2x + π / 6) - 2 y = sin (π / 3-2x) y = Tan (π / 3-2x)

y=sin(2x+π/3)
Period T = π Max 1, min - 1,
Increasing interval [K π - 5 π / 12, K π + π / 12]
Decreasing interval [K π + π / 12, K π + 7 π / 12] K ∈ Z
y=2sin（2x+π/6）-2
Period T = π Max 0, min - 4,
Increasing interval [K π - π / 3, K π + π / 6]
Decreasing interval [K π + π / 6, K π + 2 π / 3] K ∈ Z
y=sin（π/3-2x）
Period T = π Max 1, min - 1,
Minus interval [K π - π / 12, K π + 5 π / 12]
Increasing interval [K π + 5 π / 12, K π + 11 π / 12] K ∈ Z
y=tan（π/3-2x）
The period T = π / 2 has no maximum value,
Decreasing interval (K π / 2 - π / 12, K π / 2 + 5 π / 12) k ∈ Z

The period of the function y = 2Sin (2x-2 Pai / 3) is_ The amplitude is_

Because the period of y = sin (ω x + φ) is t = 2 π / | ω| and the amplitude is | a |,
So the amplitude of y = 2Sin (2x-2 π / 3) is 2 and the period is 2 π / 2 = π

The distance between the two symmetry axes of the image of the function y = root 2Sin (2x / 5 + π / 4) is

The symmetry axis of the image of the function y = 2Sin (2x / 5 + π / 4) is the same as that of the image of the function y = 2Sin (2x / 5 + π / 4),
The period of y = 2Sin (2x / 5 + π / 4) is t = 2 π / (2 / 5) = 5 π
The distance between two adjacent symmetry axes is t / 2 = 5 π / 2
The distance between the two symmetry axes of an image with y = Radix 2Sin (2x / 5 + π / 4) is 5 π / 2

How to change the image of the function y = root 2Sin (2x + Pai / 4) into the function y = radical 2cosx Write down the specific process I don't know how to change sin to Cos

With the help of induction formula: cosx = sin (x + π / 2)
So, you just change y = √ 2Sin (2x + π / 4) to y = √ 2Sin (x + π / 2)
Wish you happy! Hope to help you, if you do not understand, please ask, I wish you progress_ ∩)O

Find the symmetry center of the image of the function y = 2Sin (2x - π / 3)

When 2x - π / 3 = k π, K belongs to Z, y = 0
When x = k π / 2 + π / 6, K belongs to Z, y = 0
Therefore, the symmetry center of the image of the function is (K π / 2 + π / 6,0), and K belongs to Z

What is the symmetry of the image of the function f [x] = 2Sin [2x + π / 3]

2X + π / 3 = k π, x = k π / 2 - π / 6, i.e. symmetric about the center of a point (K π / 2 - π / 6,0)
2 x + π / 3 = k π + π / 2, x = k π / 2 + π / 12, i.e. the line x = k π / 2 + π / 12 is axisymmetric
The above K is an integer

If the image of function y = 2Sin (2x + φ) is shifted to the right by π / 4 units, the image obtained is symmetrical about the point (π / 3,0), Find the minimum of absolute value of φ

Shift π / 4 units to the right
y=2sin[2(x-π/4)+φ]
The center of SiNx symmetry is the intersection of SiNx and X axes
So the center of symmetry is on the function image
2sin[2(π/3-π/4)+φ]=0
π/6+φ=0
φ=-π/6

On whom is the image of the function y = 2Sin (2x - π \ \ 3) symmetric?

Presymmetry
2x-π/3=kπ+π/2
x=kπ/2+5π/12
Point symmetry 2x - π / 3 = k π
x=kπ/2+π/6

Given the function f (x) = the square of root 1-x / the absolute value of X + 2 - 2, judge the parity of the function

f(x)=√(1-x^2)/[|x+2|-2]
because:
In √ (1-x ^ 2), the square root number is not negative
There are:
1-x^2>=0
Get: - 1=

Let f (x) = (x ^ 2) + {X-2} - 1, judge the parity of the function (the {} in the title is absolute value, Find the minimum value of the function again

Non odd non even
When x > = 2, f (x) = x ^ 2 + x-3
When x = 2, there is a minimum value f (2) = 3
When x < 2, f (x) = x ^ 2-x + 1
When x = 1 / 2, there is a minimum value f (1 / 2) = 3 / 4
In conclusion, f (x) min = f (1 / 2) = 3 / 4