In order to get the image of the function y = cos2x, we can move the image of y = sin (2x - π / 6)

In order to get the image of the function y = cos2x, we can move the image of y = sin (2x - π / 6)

In order to get the image of the function y = cos2x,
The image is y = cos 2x = sin (2x + π / 2)
Let P (x, y) be translated by vector a (h, K) on the image of y = sin (2x - π / 6) to obtain the image of function y = cos2x, that is, the image of y = sin (2x + π / 2)
When the point P is translated by vector a (h, K), the point P '(X'Y') is obtained
X'=X+h,y'=y+k.
There is,
X=X'-h,y=y'-k.
By substituting X and Y into y = sin (2x - π / 6), we obtain that,
y'-k=sin[2(x'-h)-π/6]=sin(2x'-2h-π/6),
That is, y '= sin (2x' - 2H - π / 6) + K
If the images of Y '= sin (2x' - 2H - π / 6) + K. and y = sin (2x + π / 2) are identical, then there are
-2h-π/6=π/2, h=-π/3.
0=k.
That is to say, after translation by vector a (- π / 3,0), the
Y = cos2x image
Or by the translation of vector a (2k π - π / 3,0)
Y = cos2x image. (k belongs to Z)

In order to get the image of function y = cos (x + U / 4), how to move the image of y = SiNx?

y=cos(x+π/4)
=sin(x+π/4+π/2)
=sin(x+3π/4)
Therefore, the image of y = cos (x + π / 4) is obtained by moving the image of y = SiNx to the left by 3 π / 4 units

To get the image of the function y = SiNx, we need to change the function y = cos (x - π) 3) Image of () A. Shift π to the right 6 units B. Shift π to the right 3 units C. Shift π to the left 3 units D. Shift π to the left 6 units

Because of the function y = SiNx = cos (x - π)
2) So we only need to change the function y = cos (x − π)
3) Right shift π
You can get 6
The function y = cos (x - π)
2) The image of,
Therefore, a

The image of function y = sin (x - π / 3) only needs to translate the image of y = cos The teacher said that cos (x + π / 2) is used to get the result. Why can't we get it by subtracting π / 2 Cosine is even function`

This has nothing to do with even functions,
It's the induction formula
cos(x+π/2)=-sinx
cos(x-π/2)=sinx
So sin (x - π / 3) = cos (x - π / 3 - π / 2) = cos (X-5 π / 6)
therefore
The image of the function y = sin (x - π / 3) only needs to shift the image of y = cosx to the right by 5 π / 6 units

In order to get the image of the function y = cos (x / 2 - π / 4), we only need to transform the image of y = sin (x / 2) My practice is like this: because COS is a even function, I convert y = cos (x / 2 - π / 4) to y = cos (π / 4-x / 2) = cos1 / 2 (π / 2-x-x), and according to the induction formula, convert y = sin (x / 2) into y = cos (π / 2-x / 2) = cos1 / 2 (π - x) according to the induction formula, that is, to get cos1 / 2 (π / 2-x-x), that is, to get cos1 / 2 (π / 2-x), only need to do cos1 / 2 (π - x)? How about cos1 / 2 (π - x)? That should be right translation π / 2? Ah, why is the answer is left left left left left left left left left left left left left left? Why is the what about it? I've read it many times and asked for answers

Your solution is a little complicated. There is a problem, that is, when translating, you should pay attention to the fact that x is in front, so it should be converted into cos1 / 2 (x - π / 2), cos1 / 2 (x - π), so according to the left + right - rule, it is to shift π / 2 to the left!

Let f (x) = sin (Wx - π / 6) - 2 (COS Λ 2) (w / 2) x + 1 (W > 0), the straight line y = √ 3 and the function y = f (x) image Let f (x) = sin (Wx - π / 6) - 2 (COS Λ 2) (w / 2) x + 1 (W > 0), the distance between the two adjacent intersection points of the line y = √ 3 and the image of function y = f (x) is π, and the value of W is calculated

f(x)=sin(ωx-π/6)-2cos²(ω/2)x+1
=(√3/2)sinωx-(1/2)cosωx-cosωx
=(√3/2)sinωx-(3/2)cosωx
=√3[(1/2)sinωx-(√3/2)cosωx]
=√3sin(ωx-π/3),
Because the intersection point of the line y = √ 3 and the function y = √ 3sin (ω X - π / 3) is the vertex,
The distance π between two adjacent vertices is a period,
So ω = 2 π / T = 2 π / π = 2. (ω > 0)

What is the difference between y = sin (x + 1 / 2) and y = SiNx + 1 / 2? What is the difference between images? What is the period?

Y = sin (x + 1 / 2) is to shift y = SiNx to the left by 1 / 2 units, while y = SiNx + 1 / 2 is to shift y = SiNx up 1 / 2 units

How to change the image of y = SiNx into the image of y = sin (2x + π / 4), two answers

(1) First zoom, then translate y = SiNx, the ordinate of all points in the image remains unchanged, and the abscissa changes to the original 1 / 2, and then y = sin2 (x + π / 8) = sin (2x + π / 4) (2) shift the image to the left by π / 8 units to obtain y = sin2 (x + π / 8) = sin (2x + π / 4) (2)

How does the image of function f (x) = sin (π X / 2 - π / 4) be transformed from the image of function y = SiNx?

There are two methods: (1) first shift and then stretch y = SiNx, shift π / 4 units to the right, get y = sin (x - π / 4), then change the ordinate of all points unchanged and the abscissa to 2 / π times of the original, that is to say, the image of F (x) = sin (π X / 2 - π / 4) can be obtained. (2) the ordinates of all points of y = SiNx are invariant after stretching and then translating

Geography knowledge network map volume 1, Grade 7 More details ~ ~ ~ remember: PEP ~ ~ ~ ~ kneel down for three days```

The earth's rotation around the earth's axis from west to East is called rotation, and the earth's rotation around the sun at the same time is called revolution