Is y = 1 + SiNx called sine function image

Is y = 1 + SiNx called sine function image

It's called a sine curve

How can the image of function y = 3sin (2x + U / 3) be changed from the image of y = SiNx?

The image of the function y = SiNx is shifted to the left by π 3,
The graph of the function y = sin (x + π 3) is obtained,
The abscissa of all the points in the image is changed to 1 / 2 times of the original (the ordinate is unchanged),
Lengthen the ordinate to 3 times the original

How to transform the image of function y = SiNx to get the image of function f (x) = 3sin (1 / 2x + π / 4) - 1? Write two ways to change

1. Firstly, f (x) is expanded 3 times along Y-axis to 3sin (x); then 3sin (x) is shifted down one unit to 3f (x) - 1, then the function is stretched twice along x-axis to 3sin (1 / 2x) - 1; then, 3sin (1 / 2x) - 1 is shifted to the left by π / 4.2, sin (x) is shifted left by π / 2 units along X axis to become sin (x + π / 2)

(1) Given the function y = 3sin (2x + Pai / 3), the equation of its symmetry axis is obtained. The image of y = 3sin (2x + Pai / 3) can be obtained by the graph of y = SiNx What kind of transformation does it take?

1.y=3sin(2x+π/3) = 3 sin[ 2(x+π/6)],
2(x+π/6) = π/2 => x =π/12
=>The axis of symmetry x = k π + π / 12, or x = k π + 7 π / 12
2. For the image with y = SiNx, the amplitude is enlarged by 3 times, and π / 6 units are shifted to the left, and the X direction is compressed to 1 / 2 of the original

The function y = 3sin (2x + π) 3) The image of y = SiNx can be obtained by which of the following transformations () A. Shift π to the right 3 units. The abscissa is reduced to 1 2 times, the ordinate is expanded to 3 times of the original B. Shift π to the left 3 units. The abscissa is reduced to 1 2 times, the ordinate is expanded to 3 times of the original C. Shift π to the right Six units. The abscissa is doubled and the ordinate is reduced to 1 3 times D. Shift π to the left 6 units. The abscissa is reduced to 1 2 times, the ordinate is reduced to 1 3 times

Shift the image from y = SiNx to the left π
The function y = sin (x + π) is obtained
3）
Then the abscissa is reduced to the original 1
The function y = sin (2x + π) is obtained by changing the ordinate
3）
Then the abscissa remains unchanged and the ordinate becomes 3 times of the original, and y = 3sin (2x + π) is obtained
3）
Therefore, B is selected

The graph of the function y = x / (SiNx), X ∈ (- π, 0) ∪ (0, π) may be A reasonable explanation is required

Since y = x / sin (x) is an even function, it is only necessary to draw the image of (0, π), and then symmetrical along the y-axis, that is, the image of another interval

Limx -- > the limit of a (SiNx Sina) / (x-a)

The derivative can be used to find the limit~
Let f (x) = SiNx, then f (a) = Sina
So f '(a) = LIM (x - > A) f (x) - f (a) / x-a
Because f '(a) = (Sina)' = cosa
So the original formula = cosa

How to find the function limit x → alimsinx Sina / x-a

limsinx-sina/x-a=limcos(x+a)/2*sin(x-a)/2/[(x-a)/2]=cosa

The limit problem limx → a (SiNx Sina) / (x-a) =? Do not use the law of Roberta I haven't learned it

lim]2cos(x+a)/2sin(x-a)/2]/(x-a)=cosa

What is the limit of the function ((SiNx) ^ 2 - (Sina) ^ 2) / (x-a) at X - > A? (a is known) do not use the lophida rule!

lim(x->a) [ (sinx)^2-(sina)^2 ] /（x-a）=lim(x->a) [sinx+sina]{ [sinx-sina] /（x-a)}=lim(x->a) [sinx+sina]{ 2cos[(x+a)/2]sin[(x-a)/2]/（x-a)}=lim(x->a) [sinx+sina]*cos[(x+a)/2]*{sin[(x-a)/2]/[(x-a)/2]...