If the image of function y = SiNx is symmetric about y-axis, the image obtained is shifted to the left by π / 4 unit length. The analytic expression of the image obtained is ()

On Y-axis symmetry
Then y does not change and X becomes the opposite number
So y = sin (- x)
Shift π / 4 to the left
Left plus right minus
Then x becomes x + π / 4
So y = sin [- (x + π / 4)]
That is, y = - sin (x + π / 4)

For the function f (x) = SiNx + 2cosx, is there any φ∈ (0, R) such that the image of F (x + φ) is symmetric about the Y axis

If the proposition is true, then f (x + φ) + F (- x + φ) = sin (x + φ) + 2cos (x + φ) + sin (- x + φ) + 2cos (- x + φ) = 2cosxsin φ + 4cosxcos φ = 0. If x = 0 + a π (a is an integer), then f (x + φ) + F (- x + φ) = sin (x + φ) + 2cos (x + φ) + 2cos (- x + φ) = 2cosxsin φ + 4cosxcos φ = 0. When x = 0 + a π (a is an integer), there is tan φ = - 2 when x is not equal to 0 + a π (a is an integer)

Let y = If the image of 3cosx − SiNx is shifted to the left by M (M > 0) units, the resulting image is symmetric about the y-axis, then the minimum value of M is () A. π Six B. π Three C. 2π Three D. 5π Six

Let y =
3cosx−sinx=2cos（π
6 + x) to the left by M (M > 0) units,
The analytic expression of the function corresponding to the obtained image is y = 2cos (π
6+x+m），
Since the image obtained is symmetric about the y-axis, the function y = 2cos (π
6 + X + m) is even function, so the minimum value of M is 5 π
6，
Therefore, D

Let y = If the image of 3cosx − SiNx is shifted to the left by M (M > 0) units, the resulting image is symmetric about the y-axis, then the minimum value of M is () A. π Six B. π Three C. 2π Three D. 5π Six

When the image of the function y = root 3cosx + SiNx (x belongs to R) is shifted to the right by M (M > 0), the obtained image is symmetric about the y-axis, then the minimum value of M is () A15 degree B30 degree C60 degree D150 degree

Hope to adopt

If the unit n of the cosin = n is the smallest, then the unit of the image is the smallest?

y=sinx-√3cosx=2sin(x-π/3)
Because the image of y = SiNx - √ 3cosx shifts n units to the right, the image is symmetric about the y-axis
So the image of y = 2Sin (x-n - π / 3) is symmetric about the y-axis
So 2Sin (0-n - π / 3) = ± 1
So the minimum value of n is n = π / 6
If you do not understand, please hi me, I wish you a happy study!

Given the quadratic function y = x2-4x + 1, find 1) the image of this function symmetrical about the y-axis 2) the image symmetrical with the x-axis 3) the image symmetrical about the origin of the function

The vertex, the symmetry axis and the opening direction are the three elements of a parabola. To determine a parabola, we need only know these three elements. In this problem, the known function image is a parabola.

"If the images of two functions are symmetric about both the x-axis and the y-axis, then the images of the two functions are symmetrical about the origin!" is this proposition correct?

absolutely right

Is the image of the function y = SiNx / 3cos2x / 3 + cosx / 3sin2x / 3 symmetric about the x-axis, or the y-axis or the origin?

First understand the author's intention
It is obvious that f (x) ≠ f (- x) is an arbitrary acute angle for even functions of y-axis symmetry
In the case of X symmetry, X corresponds to two Y in a definition domain. Obviously, X and y are general functions, not multi valued functions
When the origin is symmetric, the function is odd. If x is an arbitrary acute angle, f (- x) = - f (x), it is an odd function
To sum up, it is the original function, and the image is symmetric about the origin

Given y = sin (cosx) ^ 2 * cos (SiNx) ^ 2, find y ' The answer is - sin 2xcos (cos2x),

y=sin(cosx)^2*cos(sinx)^2
y'=[sin(cosx)^2]'cos(sinx)^2+sin(cosx)^2*[cos(sinx)^2]'
={cos(cosx)^2 * (2cosx)*(-sinx)}cos(sinx)^2+
sin(cosx)^2*{-sin(sinx)^2 * (2sinx)*(cosx)}
=-2cosx(sinx)^2*cos(cosx)^2*cos(sinx)^2 - 2sin(cosx)^2*sin(sinx)^2*sinxcosx
=-2cosxsinx[cos(cosx)^2*cos(sinx)^2+sin(cosx)^2*sin(sinx)^2]
=-sin2xcos[(cosx)^2-(sinx)^2]
=-sin2xcos(cos2x)
If you look at it step by step, you will understand that it is a little more complicated, but as long as you follow the derivative formula of composite function, I hope it can help you

If the image of function y = SiNx is symmetric about y-axis, the image obtained is shifted to the left by π / 4 unit length. The analytic expression of the image obtained is ()