Sketching graphs of trig functions. Please help :(

Remember the general rules for transformations.

So for A, the transformation is directly affecting theta, it is being multiplied by 2/3. What sort of transformation occurs when the x values are multiplied by a factor?

Question B, there are no transformations that directly affect theta. But instead, once cos(theta) has been calculated, it is multiplied by 2 and then 1 is subtracted. What transformations occur when y is multiplied by a factor and when is has a value added/subtracted from it?

Thinking about it in this way could give you a clue about where to start....

Reply 3

OP

Original post by TheGrinningSkull

Best way to do sketches is to just substitute values at intervals into the equation and see the points you need to plot.
E.g. θ= -180, -135, -90, etc, just switching the values in the calculator and then make the plots

Yeah i think i can do that bit. But i'm just confused by the 2/3 in the first one. And 2 infront of sin and the -1 in the second.

Reply 4

OP

Original post by Ranibizumab

Remember the general rules for transformations.

So for A, the transformation is directly affecting theta, it is being multiplied by 2/3. What sort of transformation occurs when the x values are multiplied by a factor?

Question B, there are no transformations that directly affect theta. But instead, once cos(theta) has been calculated, it is multiplied by 2 and then 1 is subtracted. What transformations occur when y is multiplied by a factor and when is has a value added/subtracted from it?

Thinking about it in this way could give you a clue about where to start....

I am not sure which transformation occurs in either. If i had to guess, does it mean the theta is shrunk by 2/3? Not sure how this can be shown on the graph, or how to find the values from it. For the 2nd, did you not mean Sin?

Reply 5

TheGrinningSkull

19

Original post by Joburger

Yeah i think i can do that bit. But i'm just confused by the 2/3 in the first one. And 2 infront of sin and the -1 in the second.

They're basically equations,

like

y=x+1

or

y=\dfrac{3x^2}{2}-4

Reply 6

Original post by Joburger

I am not sure which transformation occurs in either. If i had to guess, does it mean the theta is shrunk by 2/3? Not sure how this can be shown on the graph, or how to find the values from it. For the 2nd, did you not mean Sin?

Oops, yes I meant sin!

For A, yes I think the range of values you need to find will be shrunk by 2/3, (because imagine the graph is being stretched in the x direction by a factor of 3/2)

For B, the transformations are not affecting theta directly, so the range of values you need to find the points in will be unchanged. :smile:

Reply 7

OP

Original post by Ranibizumab

Oops, yes I meant sin!

For A, yes I think the range of values you need to find will be shrunk by 2/3, (because imagine the graph is being stretched in the x direction by a factor of 3/2)

For B, the transformations are not affecting theta directly, so the range of values you need to find the points in will be unchanged. :smile:

For the graph for A do i start off drawing the normal cos graph? And then plot the values after multiplying theta by 2/3 for a new graph?

How do i draw the graph for B if theta isn't affected, and it's different to the standard sin graph?

Reply 8

Original post by TheGrinningSkull

Best way to do sketches is to just substitute values at intervals into the equation and see the points you need to plot.
E.g. θ= -180, -135, -90, etc, just switching the values in the calculator and then make the plots

You can use values in a calculator to help, but I would use this at the end as a way of double checking I was correct rather than the first step in solving to be honest :s

Reply 9

brianeverit

Original post by Joburger

So i got this problem from my tuition teacher but he's away on holiday now and i can't ask for any advice from him.

He left me with some problems to do on a paper, but having done the majority of questions, there is just 1 part which has me baffled.

Can i get some help on how to solve this question please. Would appreciate it so much:

Sketch the graphs of the following functions for the domain -180° ≦ θ ≦ 180°

A) Y= Cos 2/3 θ

B) Y= 2Sin θ - 1

It also says i need to label all the points of interest for both graphs. Can anyone help me in understanding and solving this question?

For A. Find where

\cos\frac{2}{3}\theta=0 \mathrm{\ and\ }\pm 1

You should then have no difficulty completing the sketch.
For B it is simply the basic sine curve stretched vertically by a factor of 2 and moved 1 unit downwards.

Reply 10

Original post by Joburger

For the graph for A do i start off drawing the normal cos graph? And then plot the values after multiplying theta by 2/3 for a new graph?

How do i draw the graph for B if theta isn't affected, and it's different to the standard sin graph?

Think about what the transformed graphs would look like.

If y = ax, a graph is stretched by a factor of 1/a in the x direction.
So, for y= cos 2x/3, the graph will be stretched by a factor of 3/2 in the x direction.
The values of theta will change:
0 < theta < 180
Becomes 0 < theta *2/3< 180*2/3 (I can't find less than and equal to signs on my iPad but just pretend they are there!)

So, you draw a normal cos graph....but at theta =180, the y value of the graph is cos(180*2/3)

Reply 11

OP

Original post by brianeverit

For A. Find where

\cos\frac{2}{3}\theta=0 \mathrm{\ and\ }\pm 1

You should then have no difficulty completing the sketch.
For B it is simply the basic sine curve stretched vertically by a factor of 2 and moved 1 unit downwards.

I do Cos2/3 on the calculator and it gives me 48.18. Not sure if that's right or how i am meant to do it? No idea how to sketch the graph for it. I'm a true beginner at this, so am easily confused.

Reply 12

Original post by brianeverit

For B it is simply the basic sine curve stretched vertically by a factor of 2 and moved 1 unit downwards.

Yep, so just imagine the normal sin graph.

Then all of the values are multiplied by 2. This means that the graph is stretched in the y direction, with max value of 2 and min value of -2.

Then the values all have 1 subtracted. This graph is moved down by 1 and the max value becomes 1 and the min value becomes -3. :smile:

Reply 13

OP

Original post by Ranibizumab

Think about what the transformed graphs would look like.

If y = ax, a graph is stretched by a factor of 1/a in the x direction.
So, for y= cos 2x/3, the graph will be stretched by a factor of 3/2 in the x direction.
The values of theta will change:
0 < theta < 180
Becomes 0 < theta *2/3< 180*2/3 (I can't find less than and equal to signs on my iPad but just pretend they are there!)

So, you draw a normal cos graph....but at theta =180, the y value of the graph is cos(180*2/3)

The domain in the question is -180° ≦ θ ≦ 180°. Is that different to what you just wrote, or am i getting confused. 2/3 of -180 and 180 is -120 and 120. Is that the new range?

Reply 14

Original post by Joburger

The domain in the question is -180° ≦ θ ≦ 180°. Is that different to what you just wrote, or am i getting confused. 2/3 of -180 and 180 is -120 and 120. Is that the new range?

Oops, sorry I meant –120< 2theta/3 < 120

The range will still be -180< theta < 180, but all the x values are being multiplied by 2/3.

So in a normal cos graph, cos(180) = 1.
Under this transformation, 180 is being multiplied by 2/3 (which makes it 120). So on your calculator, cos(120) will give you the new y value. So the last point on the graph is (180, cos(120)).

Reply 15

http://www.waldomaths.com/docs/TrigTrans01.pdf

There are some nice pictures here. Look at example G on the second page.
The graph shows cos(x/2).
Therefore, the cos graph has been stretched in the x direction by factor 2.
This is the similar to your question A, only the graph is being stretched by 3/2.

Example B on that link shows you the first transformation of your question B. Where the sin graph is being stretched by factor 2 in the y direction.

Reply 16

Calculating the new range of -120< 2theta/3 < 120 is helpful for finding co-ordinates of interest using a calculator, and not necessarily for sketching the graph.

However, another method would be:
1. draw a normal cos graph for -120 < x < 120. This is what the transformed graph will look like.
2. Now redraw this graph, but stretched over -180< x< 180 to get the transformed graph

Reply 17

OP

Original post by TheGrinningSkull

Best way to do sketches is to just substitute values at intervals into the equation and see the points you need to plot.
E.g. θ= -180, -135, -90, etc, just switching the values in the calculator and then make the plots

But how do i find out which values to substitute?

Reply 18

OP

Original post by Ranibizumab

http://www.waldomaths.com/docs/TrigTrans01.pdf

There are some nice pictures here. Look at example G on the second page.
The graph shows cos(x/2).
Therefore, the cos graph has been stretched in the x direction by factor 2.
This is the similar to your question A, only the graph is being stretched by 3/2.

Example B on that link shows you the first transformation of your question B. Where the sin graph is being stretched by factor 2 in the y direction.

Original post by Ranibizumab

Calculating the new range of -120< 2theta/3 < 120 is helpful for finding co-ordinates of interest using a calculator, and not necessarily for sketching the graph.

However, another method would be:
1. draw a normal cos graph for -120 < x < 120. This is what the transformed graph will look like.
2. Now redraw this graph, but stretched over -180< x< 180 to get the transformed graph

Thanks for the links. After doing the transformation, what method do i use to find the co-ordinates of interest? Is it not possible to also find these points without doing the sketch?

Reply 19

brianeverit