The image of the function y = f '(x) SiNx is shifted to the left π Then f (x) is the graph of the function y = 1-2sin2x______ .

By shifting the image of the function y = f '(x) SiNx to the left by π 4 units, y = 1-2sin2x is obtained, because f' (x + π 4) sin (x + π 4) = f '(x + π 4) × 22 (cosx + SiNx) = 1-2sin2x = cos2x = cos2x-sin2x  f' (x + π 4) = 2 (cosx SiNx) = 2cos (x + π 4)  f '(x) = 2cosx 

The image of the function y = f (x) · SiNx is shifted to the right π After 4 units, the symmetry transformation about the X axis is made to get the image of the function y = 1-2sin2x, then f (x) is () A. -2cosx B. 2cosx C. -2sinx D. 2sinx

∵ y = 1-2sin2x = cos2x, which is obtained by symmetric transformation about x-axis
Y = - cos2x, then shift π to the left
Four units get the function
y=-cos2（x+π
4）=sin2x，
That is, y = sin2x = f (x) · SiNx
∴f（x）=2cosx．
Therefore, B

The image of function y = SiNx is translated by vector a = (PI / 6,4) to get f, and find the function analytic formula of F

A positive abscissa indicates a shift to the right
A positive ordinate indicates an upward translation
So f = sin (x-pie / 6) + 4

If the function y = 2Sin (x + θ) is represented by the vector (π) After translation, one of its symmetry axes is x = π Then a possible value of θ is () A. 5π Twelve B. π Three C. π Six D. π Twelve

The image of the function y = 2Sin (x + θ) is expressed by vector (π)
6,2) after translation, the function y = 2Sin (x + θ + π) is obtained
6) + 2 image,
Because one of its symmetry axes is x = π
So π
4+θ+π
6=kπ+π
2，k∈Z，
When k = 0, θ = π
12. Satisfy the question
Therefore, D

If the image of function f (x) = 2Sin (2x - π / 6) is translated by vector a = (- π / 3,1), the image of function y = g (x) is obtained, then the function g (x) is in the interval The maximum value on [π / 6, π / 4] is

g(x)=1+2sin( 2(x+π/3)-π/6 )
=1+2sin( 2x+π/2 )
=1+2cos( 2x)
The minimum positive period is π
In [0, π / 2], it is a monotone decreasing interval
So g (π / 6) is the maximum on [π / 6, π / 4], which is 1 + 2cos (π / 3)
=1

Translate the image of function y = 2x ^ 2-4x + 5 according to vector a to get the image of y = 2x ^ 2, and a ⊥ B, C = (1, - 1), b * C = 4, find that B. letters A.B.C are vectors!

Y
=2x^2-4x+5
=2(x-1)^2+3
∴a=(-1,-3)
∵ C = (1, - 1), b * C = 4, a * b = 0, let B = (x, y),
∴x-y=4,-x-3y=0,
∴x=3,y=-1
That is, B = (3, - 1)

It is known that the symmetry center of the function image obtained by translating the image of function y = x / X-1 according to vector a is the origin

Function y = x / (x-1) = 1 + [1 / (x-1)]
The center of symmetry of this function is (1,1)
If the center of symmetry after translation according to vector a is the origin, then: a = (- 1, - 1)

Find the monotone interval of function y = SiNx (1 + cosx) (0 ≤ x ≤ 2 π)

Cosx (1 + COS) + SiNx (0-sinx) = cosx + cos ^ 2 (x) - Sin ^ 2 (x) = cosx + cos + cos (2x) = 2cos ^ 2 (x) + cosx-1 cosx + 2 (cosx) 2 (cosx) ^ 2-1 = 02 (cosx) ^ 2 + cosx-1 = 0 (cosx + 1) (2cosx-1 = 0 (cosx + 1) (2cosx-1) = 0cosx x = - 1 or cosx = 1 / 2x = π + 2K π or π + 2K π or x = π / 3 / 3, x = π + 2K π or x = π / 3 / 3 / 3, x = π + 2K π or x = π / 3 / 3 / 3 / 3, or x = π + + 2K π or - π / 3 + 2K π 0 ≤ X

Find the monotone interval (1) y = 1 + SiNx (2) y = - cosx writing process! ~~

(1) When 2K π - π / 2 < x < 2K π + π / 2, y = 1 + SiNx monotonically increases; when 2K π + π / 2 < x < 2K π + π / 2, y = 1 + SiNx monotonically decreases. (2) y = - cosx is opposite to y = cosx ﹤ when 2K π ﹤ x ﹤ 2K π + π, y = -

Monotone increasing interval of function y = cosx / (1-sinx) I know: Cosx = cos (x / 2) square - sin (x / 2) square 1-sinx = cos (x / 2) square - 2Sin (x / 2) cos (x / 2) + sin (x / 2) square =[cos (x / 2) - sin (x / 2)] squared So y = cos (x / 2) - sin (x / 2) squared / [cos (x / 2) - sin (x / 2)] squared =cos(x/2)+sin(x/2) / cos(x/2)-sin(x/2) =1+tan(x/2) / 1-tan(x/2) I am mainly stuck in this step: So y = cos (x / 2) - sin (x / 2) squared / [cos (x / 2) - sin (x / 2)] squared =cos(x/2)+sin(x/2) / cos(x/2)-sin(x/2) =1+tan(x/2) / 1-tan(x/2) Why can you take out the square directly? I can't take the numerical value in!

It's not taking out the square, it is
Y = cos (x / 2) square - sin (x / 2) square / [cos (x / 2) - sin (x / 2)] square
=(COS (x / 2) - sin (x / 2)) (COS (x / 2) + sin (x / 2)) / [cos (x / 2) - sin (x / 2)] squared
=Cos (x / 2) + sin (x / 2) / cos (x / 2) - sin (x / 2) reduction

The image of the function y = f '(x) SiNx is shifted to the left π Then f (x) is the graph of the function y = 1-2sin2x______ .