The image of function y = SiNx is transformed into the image of function y = 3sin (2x + π / 3) + 1 by two methods RT, in two ways

The image of function y = SiNx is transformed into the image of function y = 3sin (2x + π / 3) + 1 by two methods RT, in two ways

First shift π / 3 unit length to the left, then change the abscissa to 2 / 1 of the original, then change the ordinate to 3 times of the original, and finally move up 1 unit length
First, the abscissa is changed to 2 / 1 of the original, then π / 6 unit length is shifted to the right, then the ordinate is changed to 3 times of the original, and finally 1 unit length is shifted upward

Given the function y = 3sin (1 / 2x - π / 4), the equation of the symmetry axis of this function image is obtained

(1) Axis of symmetry
1/2x--π/4=kπ+π/2
1/2x=kπ+3π/4
Symmetry axis X = 2K π + 3 π / 2, K ∈ Z
(2) Center of symmetry
1/2x--π/4=kπ
1/2x=kπ+π/4
x=2kπ+π/2
Symmetry center (2k π + π / 2,0) k ∈ Z

What are the coordinates of all symmetry centers of the function y = 3sin (2x + π / 3)? The equation of all symmetry axes? RT. etc How to find out..

0

0

By 2x + π
6=kπ+π
2, x = k π
2+π
6（k∈Z），
Let k = 0, then x = π
6，
The equation of its symmetry axis is x = π
6，
Therefore, C

The symmetric axis equation of the function y = 3sin (2x + π / 3) is?

The function of sine function and the idea of global substitution
Let 2x + π / 3 = t, then y = 3sint. The symmetric axis equation of the sinusoidal function is that if t = π / 2 + K π, K belongs to Z
Then 2x + π / 3 = π / 2 + K π, K belongs to Z
X = - π / 12 + K π / 2, K belongs to Z
Do you understand? We need to study the symmetry axis of sine function, and then deduce the symmetry axis of sine function
I don't know how to ask questions~

In order to get the image of the function y = cosx (x + π / 3), we need only change the image of function y = SiNx_________ How to understand how to translate it to the left by 5 π / 6 units of length

The image of y = cosx can be seen as the image of y = SiNx is shifted to the left π / 2, while the image of y = cosx (x + π / 3) is obtained by shifting the image of y = cosx by π / 3, so the image of y = cosx (x + π / 3) is obtained by shifting the image of y = SiNx to the left by π / 2 + π / 3 (= 5 π / 6)

When x approaches a, find the limit of (SiNx Sina) / x-a

Method 1: SiNx sina is transformed into 2cos [(x + a) / 2] · sin [(x-a) / 2] by using the sum difference product formula, and then replaced by equivalent infinitesimal
lim(x→a) [(sinx-sina)/(x-a)]
=lim(x→a) 2cos[(x+a)/2]·sin[(x-a)/2]/(x-a)
=2cosa*lim(x→a) [sin[(x-a)/2]/(x-a)
=2cosa*(1/2)
=cosa
Method two: lophida's rule
lim(x→a) [(sinx-sina)/(x-a)]
=lim(x→a) [(sinx-sina)'/(x-a)']
=lim(x→a) cosx
=cosa

The limit of (SiNx Sina) / x-a when x tends to a

X → a, SiNx Sina → 0, (x-a) → 0
x→a,(sinx-sina)'/(x-a)'=cosx/1→cosa
So the limit is cosa
Note:
The conditions for satisfying the law of l'urbida are as follows
The denominator tends to zero or infinity at the same time

Find the limit of (SiNx Sina) / (x-a) when x tends to a Write a general idea,

The sum difference product of molecules must have sin (x-a), so it is directly converted to x-a

(SiNx Sina) / (x-a) when x → a, what is its limit? How to calculate it, As the title

If you remember correctly, use the Loby's law to get cosx / 1 = cosa