# Math Help - Phase shift??

1. ## Phase shift??

I have sketched the functions of y=sin(x) and y=3sin(x) but when i try and sketch y=3sin(x-1pi), i get a bit confused, i can disern the difference with adding or subtracting fractions of Pi from (x) and observing a slight shift but when adding/subtracting a whole Pi to (x) won't it look exaclty the same?

2. Originally Posted by bobchiba
I have sketched the functions of y=sin(x) and y=3sin(x) but when i try and sketch y=3sin(x-1pi), i get a bit confused, i can disern the difference with adding or subtracting fractions of Pi from (x) and observing a slight shift but when adding/subtracting a whole Pi to (x) won't it look exaclty the same?
it won't be exactly the same. shifting by 2pi would make it look exactly the same. we still begin at the same place, but at x = 0 for instance, instead of going up first, we go down

maybe this will help. the first part is an introduction to the concepts you'll need

http://www.mathhelpforum.com/math-he...functions.html

(if it doesn't help, say so)

3. Here's sin(x) for the first period starting at x = 0

4. Here's 3sin(x). note that the amplitude changes

5. Here is 3sin(x - pi). we have shifted pi units to the right. if you trace the graph backwards to x = 0, you notice we dip before rising, that is, we actually resemble a -3sin(x) graph

6. I'll be honest, I got a bit lost in that other post, this is what I have for
y=sin(2X), but i can't visualize it -Pi .....(y=sin(2x-Pi))
If u could maybe edit my pic and re-post it up would be good?

7. Originally Posted by bobchiba
I'll be honest, I got a bit lost in that other post, this is what I have for
y=sin(2X), but i can't visualize it -Pi .....(y=sin(2x-Pi))
If u could maybe edit my pic and re-post it up would be good?
y = sin(2x - pi)
=> y = sin2(x - pi/2)

the phase shift is how many units we move to the left or right. when we have the graph in that form, the phase shift is just whatever number we are subtracting from x (we change the sign). here, the phase shift is pi/2. it's positive, so we move to the right. basically what we do is we take the graph and drag it to the right pi/2 units. so where we started at zero, we now start at pi/2. the point that was at 1 is now at 1 + pi/2, the point that was at pi/2 is now at pi, etc... so what you are doing here, is taking the y = sin(2x) graph, and shifting it pi/2 units to the right. so the graph will look the same, just shifted a bit.

if you multiply by a constant, say 3, it just stretches the graph vertically. the amplitude becomes 3. so we go 3 units above the resting point, and 3 units below, as you drew.

got it?

8. I will post a graph of y = 3sin(2x - pi). the red dot that you see on the graph, will be the dot that was originally at the origin. you will notice that it shifted pi/2 units, so now it is at (pi/2, 0) where it was first at (0,0)

Here it is

9. So you see that in the first graph (y = 3sin(2x)) we started at the origin and went up. now when we start at the origin ( in y = 3sin(2x - pi)) we go down first at the origin. since we shifted pi/2 units to the right. the red dot you see represents the point that was originally at the origin in the graph you drew

10. By the way, sorry Jhevon etc, I started the post with 3sinx-Pi ..... I meant to write 3sin2x-Pi like in my attached post....sorry.

11. Originally Posted by bobchiba
By the way, sorry Jhevon etc, I started the post with 3sinx-Pi ..... I meant to write 3sin2x-Pi like in my attached post....sorry.
that's fine. you have the answer now anyway. the question is, do you get how to get to the answer? you should read the first part of the post i suggested again, very slowly. don't let the letters confuse you, the just represent numbers. you can match them up with the formula you are trying to graph and it makes life simple

12. Ok, thanks alot I get it now, when you changed it to y=sin2(X-Pi/2) helped alot.

I presume the period, amplitude and A'frequency are the same for this phase in its own right.