Announcements
#21
(Original post by brianeverit)
The points of interest will be, intersections with the axes and maximum and minimum points
For example
How do you work out 135?
0
5 years ago
#22
(Original post by Joburger)
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?
Points of intersection, max and min points. For example,
Cos2theta/3 = 0
Arcos(0) = 2theta/3 = 90
So theta = 90*3/2
Then use the periodicity if the graph to find the other values of theta that equal 0 in the region -120< x <120

Yes it is possible to do without sketching the graph, but I think whilst you are still learning you should sketch the graphs each time. Will help you understand transformations and also will help you see how many points there are to find.

For example, would you have a rough idea of how many maximum and minimum points to expect for cos(3x), cos(0.5x) or cos(2x) without sketching the graph?
Or could you easily sketch the graph of y= 3cos(3x) + 3 ?
If you need the practice it would be a good idea to draw them. Eventually it will become second nature
0
#23
(Original post by Ranibizumab)
Points of intersection, max and min points. For example,
Cos2theta/3 = 0
Arcos(0) = 2theta/3 = 90
So theta = 90*3/2
Then use the periodicity if the graph to find the other values of theta that equal 0 in the region -120< x <120

Yes it is possible to do without sketching the graph, but I think whilst you are still learning you should sketch the graphs each time. Will help you understand transformations and also will help you see how many points there are to find.

For example, would you have a rough idea of how many maximum and minimum points to expect for cos(3x), cos(0.5x) or cos(2x) without sketching the graph?
Or could you easily sketch the graph of y= 3cos(3x) + 3 ?
If you need the practice it would be a good idea to draw them. Eventually it will become second nature
Thank you again. Yes you are right, i need more practice drawing them. Even now i am still confused about how to do it. When you say the periodicity of the graph, is this for the new graph, after "2/3" has affected it? I am not sure i have it completely right. Have you worked out the values for q A? Would be great if you could post them so i can see how it's done working backwards, and then i'd be more confident doing B too. Thanks for the help thus far though
0
5 years ago
#24
(Original post by Joburger)
But how do i find out which values to substitute?
It's not like a set rule. Just plug in the limits (which in the question says -180 to 180.

Then at intervals, e.g. 90 degrees. Maybe more e.g. every 45 degrees just to see the nature of the curve and in which direction they curve.
0
#25
Bump. Still confused with how the graph for question B will look like and how to draw it, and find the values. Any help?
0
#26
Any ideas on the answers to A and B?
0
5 years ago
#27
(Original post by Joburger)
Thank you again. Yes you are right, i need more practice drawing them. Even now i am still confused about how to do it. When you say the periodicity of the graph, is this for the new graph, after "2/3" has affected it? I am not sure i have it completely right. Have you worked out the values for q A? Would be great if you could post them so i can see how it's done working backwards, and then i'd be more confident doing B too. Thanks for the help thus far though
Sorry, I was busy for the past few hours! To be honest question B is the easier one, it would be a better one to start with

Ok so by periodicity I mean that the graph repeats itself every few degrees.
Think of normal cos. it crosses the y axis at 90, 270, -90 and -270.
+/- 270 is outside your range, and as the graph has been stretched by a factor of 3/2, these areas will have not moved into the range (if the graph had been squashed instead, then we may need to consider values from outside the range)

So I've already mentioned earlier how to show that 2theta/3 = 90, which by rearranging gives theta = 135.
This can also be used for -90, which gives theta = -135.
So you have points of (135, 0) and (-135, 0).
0
#28
(Original post by Ranibizumab)
Sorry, I was busy for the past few hours! To be honest question B is the easier one, it would be a better one to start with

Ok so by periodicity I mean that the graph repeats itself every few degrees.
Think of normal cos. it crosses the y axis at 90, 270, -90 and -270.
+/- 270 is outside your range, and as the graph has been stretched by a factor of 3/2, these areas will have not moved into the range (if the graph had been squashed instead, then we may need to consider values from outside the range)

So I've already mentioned earlier how to show that 2theta/3 = 90, which by rearranging gives theta = 135.
This can also be used for -90, which gives theta = -135.
So you have points of (135, 0) and (-135, 0).
Thanks very much. So it's just 4 values for A? I do understand how you worked that out now. So thank you. Funnily enough, Question B seems harder for me. Even though theta isn't affected, i'm still confused at what the correct values are and how to find them
0
5 years ago
#29
As for max and min points, think of where they are in normal cos.

Normal cos has max points (0,1), (360, 1) (-360, 1)
And min points of (180, -1) and (-180, -1)

Under the transformation, the graph is getting stretched, so the points at +/-360 will not have been moved into the range. Likewise, the minimum points have probably moved out of the range.

The point (0,1) will be the same under the transformation.

The new minimum point won't have reached -1 under the transformation, but the lowest point will be at x coordinates +/-180
So, the min points will be (180, cos(2*180/3)) and (-180, cos(2*-180/3))
0
5 years ago
#30
(Original post by Joburger)
Thanks very much. So it's just 4 values for A? I do understand how you worked that out now. So thank you. Funnily enough, Question B seems harder for me. Even though theta isn't affected, i'm still confused at what the correct values are and how to find them
Well, there are two value within your range that you have been given. Normal cos of -360 < x < 360 would contain 4 points that cross the x axis. However your range is restricted to -180 < x 180.
You also have to consider the transformation, which could stretch or squash some values out of or into the range.

Why it's important to check the range!

It can be useful to use online graphic calculators as they will show you the transformed graph. Just zoom in and see where the graph reaches the range required. Anything outside of this range can be ignored.
0
5 years ago
#31
Question B is easy! Nothing is happening to the x values, so you don't have to worry whether the points of interest are in the range or not.

Just think of the sin graph, draw it from -180 to 180.

Max point at: (90, 1)
Min point at (-90,-1)

Under the transformation, the x values aren't being changed, but y is.
Transformation is sinx *2 = 2sinx
Then 2sinx - 1
So, do this to each y value.
1*2 - 1 = 1 = new max value -> (90, 1)
-1*2 -1 = -3 = new min value -> (-90, -3)
0
5 years ago
#32
For question B, the points where the graph crosses the axis will be different.

Best way to work this out is algebraically:

2sinx -1 = 0
2sinx = 1
Sinx = 1/2
Arcsin(1/2) = x

X = 30

Now the periodicity of the graph can be used to work out where the other points of axes intersection are.
As the graph has shifted downwards, only point between 0 and 180 will cross the x axis.

The graph is symmetrical, so if it crosses the x axis at (30,0), then it also crosses at (150,0) under this transformation. (0 to 30 is the same distance as 180 - 30)
0
#33
(Original post by Ranibizumab)
For question B, the points where the graph crosses the axis will be different.

Best way to work this out is algebraically:

2sinx -1 = 0
2sinx = 1
Sinx = 1/2
Arcsin(1/2) = x

X = 30

Now the periodicity of the graph can be used to work out where the other points of axes intersection are.
As the graph has shifted downwards, only point between 0 and 180 will cross the x axis.

The graph is symmetrical, so if it crosses the x axis at (30,0), then it also crosses at (150,0) under this transformation. (0 to 30 is the same distance as 180 - 30)
Wish i could rep you more than once. Thanks very much for the explanations. I understand it alot better now, and will get more practice of these questions to try and fully master it. Great help!
0
5 years ago
#34
No problem, I would keep sketching the graphs, and use an online graphics calculator to double check!

Once you can see what's happening it becomes a lot easier
0
5 years ago
#35
(Original post by Joburger)
How do you work out 135?
0
X

new posts
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• Arts University Bournemouth
Art and Design Foundation Diploma Further education
Sat, 25 May '19
• SOAS University of London
Wed, 29 May '19
• University of Exeter
Thu, 30 May '19

### Poll

Join the discussion

Loved the paper - Feeling positive (104)
19.01%
The paper was reasonable (275)
50.27%
Not feeling great about that exam... (109)
19.93%
It was TERRIBLE (59)
10.79%