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Trig - Sketching a graphy

This questions is from the non calculator SQA examination 2005. I'm struggling a bit on 10b) and wondered if anybody could help me out. I know I do have a tiny little red gem in the top right hand corner there but don't let that put you off :wink: I'm sitting on -1 after I started a controversial thread :frown: :tongue:

This is the question:

2005, P1, Q10.jpg

The answer to a) is 2sin(xπ3)2sin \left( x - \frac\pi{3} \right)

And this is what I attempted to do for b):
10b solution.jpg

It seems very complicated though :frown:

EDIT: The title is meant to say graph not graphy, was a bit rushed. But you must admit graphy does sound a bit cooler :tongue:

EDIT2: How can you neg me for this? :frown: Just wanting some help :frown:
(edited 11 years ago)
Reply 1
The most simple way to do this:

Draw the graph of sinx\sin x and then use transformations to get you to the graph of 2sin(xπ3)+32sin \left( x - \frac\pi{3} \right) +3

Does that make any sense or should I explain further?
Reply 2
Original post by notnek
The most simple way to do this:

Draw the graph of sinx\sin x and then use transformations to get you to the graph of 2sin(xπ3)+32sin \left( x - \frac\pi{3} \right) +3

Does that make any sense or should I explain further?


Thanks for replying, is there any chance you could explain a bit further? :tongue:
Reply 3
Original post by OpenArms
Thanks for replying, is there any chance you could explain a bit further? :tongue:

Sure. Do you remember learning about transformations of functions e.g. things like f(x)-->f(x+3), and how you can draw them on a graph?

Start with the graph of sin(xπ3)\sin\left(x-\frac{\pi}{3}\right).

Let f(x)=sinxf(x)=\sin x then f(xπ3)=sin(xπ3)f(x-\frac{\pi}{3})=\sin\left(x-\frac{\pi}{3}\right).

How do you draw the graph of f(x-a) using the graph of f(x)?

Next, notice that 2sin(xπ3)=2f(xπ3)2\sin\left(x-\frac{\pi}{3}\right) = 2f(x-\frac{\pi}{3})
(edited 11 years ago)
Reply 4
Original post by notnek
Sure. Do you remember learning about transformations of functions e.g. things like f(x)-->f(x+3), and how you can draw them on a graph?

Start with the graph of sin(xπ3)\sin\left(x-\frac{\pi}{3}\right).

Let f(x)=sinxf(x)=\sin x then f(xπ3)=sin(xπ3)f(x-\frac{\pi}{3})=\sin\left(x-\frac{\pi}{3}\right).

How do you draw the graph of f(x-a) using the graph of f(x)?


you would move it to the right if a is + and the left if a is -?
Reply 5
Original post by OpenArms
you would move it to the right if a is + and the left if a is -?

You've got it the wrong way round :smile:.

See the edit in my previous post to see what to do next.
Reply 6
Original post by notnek
You've got it the wrong way round :smile:.

See the edit in my previous post to see what to do next.


Graphs.jpg
(edited 11 years ago)
Reply 7
Original post by OpenArms
Graphs.jpg


You're correct in how you've drawn sin(xπ3)\sin(x-\frac{\pi}{3}).

Where does sinx\sin x cross the x-axis? For every point where it crosses, sin(xπ3)\sin(x-\frac{\pi}{3}) will cross π3\frac{\pi}{3} units to the right.

For the next part, you're stretching along the wrong axis. You may need to look back in your textbook to function transformations.
Reply 8
Original post by notnek
You're correct in how you've drawn sin(xπ3)\sin(x-\frac{\pi}{3}).

Where does sinx\sin x cross the x-axis? For every point where it crosses, sin(xπ3)\sin(x-\frac{\pi}{3}) will cross π3\frac{\pi}{3} units to the right.

For the next part, you're stretching along the wrong axis. You may need to look back in your textbook to function transformations.



So the graph is stretched to double upwards and has shifted to the right by π3 \frac\pi{3} ?

Graph1.jpg
(edited 11 years ago)
Reply 9
Original post by OpenArms
So the graph is stretched to double upwards and has shifted to the right by π3 \frac\pi{3} ?

Graph1.jpg

I'm not sure why the maximum of your graph has become 5. Did you draw sinx\sin x correctly? The graph of sinx\sin x is between -1 and 1 with the first maximum (starting from 0 and going right) at (π2,1)(\frac{\pi}{2},1).

The first transformation moves the maximum π3 \frac{\pi}{3} units to the right so it becomes (5π6,1)(\frac{5\pi}{6},1).

The second transformation stretches the graph along the y-axis by a factor of 2 (i.e. y is multiplied by 2) so the maximum becomes (5π6,2)(\frac{5\pi}{6},2).
(edited 11 years ago)
Reply 10
Original post by notnek
I'm not sure why the maximum of your graph has become 5. Did you draw sinx\sin x correctly? The graph of sinx\sin x is between -1 and 1 with the first maximum (starting from 0 and going right) at (π2,1)(\frac{\pi}{2},1).

The first transformation moves the maximum π3 \frac{\pi}{3} units to the right so it becomes
Unparseable latex formula:

(\frac{5\pi}{6},1}

.

The second transformation stretches the graph along the y-axis by a factor of 2 (i.e. y is multiplied by 2) so the maximum becomes (5π6,2)(\frac{5\pi}{6},2).


The solutions said it was 5 so I tried to work my way back from the solutions to create a method. I'm finding the solutions very confusing. I have attached it for you.

Markers Solution.jpg
Reply 11
Original post by OpenArms
The solutions said it was 5 so I tried to work my way back from the solutions to create a method. I'm finding the solutions very confusing. I have attached it for you.

Markers Solution.jpg


The maximum becomes 5 if you draw 2sin(xπ3)+32\sin(x-\frac{\pi}{3})+3 (which is where all this working is leading to) but I thought the graph you drew was of 2sin(xπ3)2\sin(x-\frac{\pi}{3}) which is why I questioned your maximum.

To find the y-intercept of the final graph, plug in x=0x=0 into 2sin(xπ3)+32\sin(x-\frac{\pi}{3})+3.
(edited 11 years ago)
Reply 12
Original post by notnek
The maximum becomes 5 if you draw 2sin(xπ3)+32\sin(x-\frac{\pi}{3})+3 (which is where all this working is leading to) but I thought the graph you drew was of 2sin(xπ3)2\sin(x-\frac{\pi}{3}) which is why I questioned your maximum.

To find the y-intercept of the final graph, plug in x=0x=0 into 2sin(xπ3)+32\sin(x-\frac{\pi}{3})+3.


Apologies for the confusion.

Here is the graph of 2sin(xπ3)2\sin(x-\frac{\pi}{3})
2sinx(x-pi3).jpg

and here is the graph of 2sin(xπ3)+32\sin(x-\frac{\pi}{3})+3
2sinx(x-pi3) + 3.jpg
plus some scribbling's on how to get the y value.
Reply 13
Original post by OpenArms
Apologies for the confusion.

Here is the graph of 2sin(xπ3)2\sin(x-\frac{\pi}{3})
2sinx(x-pi3).jpg

and here is the graph of 2sin(xπ3)+32\sin(x-\frac{\pi}{3})+3
2sinx(x-pi3) + 3.jpg
plus some scribbling's on how to get the y value.


Since sin is an odd function, sin(π3)=sin(π3)sin(-\frac{\pi}{3}) = -sin(\frac{\pi}{3}).

sin(π3)\sin(\frac{\pi}{3}) is something that you're expected to know the value of at C4. Maybe you're more familiar in terms of degrees: sin(60)\sin(60)?
Reply 14
Original post by notnek
Since sin is an odd function, sin(π3)=sin(π3)sin(-\frac{\pi}{3}) = -sin(\frac{\pi}{3}).

sin(π3)\sin(\frac{\pi}{3}) is something that you're expected to know the value of at C4. Maybe you're more familiar in terms of degrees: sin(60)\sin(60)?


Well I do the scottish system but I do know my exact values :tongue:

sin(60)=32\sin(60) = \frac{\sqrt{3}}{2}
Reply 15
Original post by OpenArms
Well I do the scottish system but I do know my exact values :tongue:

sin(60)=32\sin(60) = \frac{\sqrt{3}}{2}

Sorry, I'm getting my threads mixed up.

So have you completed the question now?
Reply 16
Original post by notnek
Sorry, I'm getting my threads mixed up.

So have you completed the question now?



2sin(π3)+32sin(-\frac{\pi}{3}) + 3

=2sin(π3)+3= -2sin(\frac{\pi}{3})+ 3

=2(32)+3= -2(\frac{\sqrt{3}}{2}) + 3

=232+3= -\frac{2\sqrt{3}}{2} + 3

=3+3= -\sqrt{3} + 3

=33= 3 - \sqrt{3}

Is this correct?
Reply 17
Original post by OpenArms
2sin(π3)+32sin(-\frac{\pi}{3}) + 3

=2sin(π3)+3= -2sin(\frac{\pi}{3})+ 3

=2(32)+3= -2(\frac{\sqrt{3}}{2}) + 3

=232+3= -\frac{2\sqrt{3}}{2} + 3

=3+3= -\sqrt{3} + 3

=33= 3 - \sqrt{3}

Is this correct?


Yes.

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