The monotone interval of the function f (x) = sin (2x + 45 °) + cos (2x + 45 °) is obtained, and the monotone interval is? The axis of symmetry is?

The monotone interval of the function f (x) = sin (2x + 45 °) + cos (2x + 45 °) is obtained, and the monotone interval is? The axis of symmetry is?

f(x)=√2sin(2x+45°+45°)=√2cos2x
The period is π
The monotone increasing interval is (K π - π / 2, K π)
The monotone decreasing interval is (K π, K π + π / 2)
The axis of symmetry is x = k π / 2
Here K is any integer

It is known that f (x) = cos 2 (x + π / 3) - 1 / 2 g (x) = 1 / 2Sin (2x + 2 π / 3) to find the f (x) of G (x) by how to transform And H (x) = f (x) - G (x) Finding the maximum value of H (x) and the corresponding x

F (x) = cos 2 (x + π / 3) - 1 / 2 = (1 / 2) · [2cos 2 (x + π / 3) - 1] = (1 / 2) · cos (2x + 2 π / 3)

Given the function f (x) = 2Sin ^ 2x + cos (2x-3 π) - 1, find the value range of the function on {12}, 2}

f(x)=2sin^2x+cos(2x-π/3)-1
=(2sin^2x-1)+cos(2x-π/3)
=-cos2x+1/2*cos2x+√3/2*sin2x
=√3/2*sin2x -1/2*cos2x
=sin(2x-π/6)
∵ x belongs to [π / 12, π / 2]
∴π/6≤2x≤π
That is 0 ≤ 2x - π / 6 ≤ 5 π / 6
∴0≤sin(2x-π/6)≤1
That is, the value range of function f (x) is [0,1]

Given the function f (x) = cos (2x - π / 3) + 2Sin (x - π / 4), find the value range of function on the interval [0, π / 2] Finding monotone interval of function Wrong number... F (x) = cos (2x - π / 3) + 2Sin (x - π / 4) sin (x + π / 4)

f(x)＝cos(2x-π/3)+2sin(x-π/4)sin（x+π/4）＝cos(2x－π/3)＋2sin(x－π/4)cos[π/2－(x＋π/4)]
＝cos(2x－π/3)＋2sin(x－π/4)cos(x－π/4)＝cos(2x－π/3)＋sin(2x－π/2)
＝cos(2x－π/3)＋cos2x＝2cos(2x－π/6)cosπ/6＝√3cos(2x－π/6)
∵x∈[0,π/2] ∴2x－π/6∈[﹣π/6,5π/6] ∴f(x)∈[﹣3/2,√3]
When 2x - π / 6 ∈ [2K π, 2K π + π], i.e., X ∈ [K π + π / 12, K π + 7 π / 12], it increases monotonically
When 2x - π / 6 ∈ [2K π + π, 2K π + 2 π], i.e., X ∈ [K π + 7 π / 12, K π + 13 π / 12], it decreases monotonically

Given the function f (x) = 2Sin (2x - π / 3) - 1 (1), find the maximum value of the function and the minimum value (2) if x ∈ [- π / 2, π / 2] is the increasing interval of the function (3) How can the image of the function be transformed from the image of the function y = sin2x?

(1) When 2x - π / 3 = 2K π + π / 2, that is, x = k π + 5 π / 12, the maximum value of F (x) is 1
When 2x - π / 3 = 2K π - π / 2, that is, x = k π - π / 12, the minimum value of F (x) is - 3
(2)∵x∈[﹣π/2,π/2] ∴2x－π/3∈[﹣4π/3,2π/3]
When 2x - π / 3 ∈ [- π / 2, π / 2], that is, X ∈ [- π / 12,5 π / 12], the function increases monotonically
When x ∈ [- π / 2, π / 2], the increasing interval of the function is: [- π / 12,5 π / 12]
(3)∵f(x)＝2sin(2x－π/3)－1＝2sin[2(x－π/6)]－1
The image of the function can be shifted by π / 6 units to the right along the x-axis, and then 1 unit down the y-axis

Let the maximum value of the function f (x) = 2Sin (2x + π \ \ 6) be m, and find the monotone increasing interval of the summation function; If 10 unequal positive numbers satisfy f (XI) = m and Xi

f(x)max=2=M
When 2K π - π / 2 ≤ 2x + π, 6 ≤ 2K π + π / 2
In other words, f (x) increases when k π - π / 3 ≤ 2x + π / 6 ≤ K π + π / 6
So the monotone increasing interval of F (x) is [K π - π / 3, K π + π / 6], K ∈ Z
The period of the function is π
Because Xi is 10 mutually unequal positive numbers, and f (XI) = M = f (π / 6)
So xi = π / 6, π / 6 + π, π / 6 + 2 π
So X1 + x2 + X3. + X10 = (1 + 2 + 3 +... + 9) π + π / 6 × 10 = 140 π / 3

The monotone increasing interval of the function f (x) = SiNx + cosx is__

f(x)=sinx+cosx
=√2(√2/2*sinx+√2/2cosx)
=√2(sinxcosπ/4+cosxsinπ/4)
=√2sin(x+π/4)
If increasing, 2K π - π / 2

Is the monotone decreasing interval of the function y = 2Sin (SiNx + cosx)?

Let f (x) = 2Sin (g (x)); PI = 3.14
g(x)=sinx+cosx=2^0.5sin(x+pi/4);
Then f (x) is the outer function and G (x) is the inner function. Remember that if f (x) and G (x) increase together, they will increase; if they decrease, they will decrease; if they increase and decrease, they will decrease
It's almost out of the light. Do it yourself

Primitive function of SiNx / cos ^ 2x

The original function of F (x) = SiNx / cos? X is f (x) integrating X: ∫ f (x) = ∫ SiNx / cos? X DX = ∫ - 1 / cos? X dcosx is used for dcosx = - SiNx = 1 / cosx + C for D (1 / x) = - 1 / x? Or D (1 / cos x) = - 1

Is the image of the function y = 2cos (2x + π / 3) symmetric about π / 12? Is the expression y = cos (SiNx) (x ∈ R) even? The answer is that the first sentence is wrong and the second sentence is right. It is impossible. At least the first sentence must be right Isn't it? The first is about x = π / 12 symmetry

If x = π / 12, y = 0, it is not satisfied
The function y = cos (SiNx) (x ∈ R) is an even function