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Trig - Sketching a graphy

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    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 I'm sitting on -1 after I started a controversial thread

    This is the question:

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

    And this is what I attempted to do for b):
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    It seems very complicated though

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

    EDIT2: How can you neg me for this? Just wanting some help
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    The most simple way to do this:

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

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

    Draw the graph of \sin x and then use transformations to get you to the graph of 2sin \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?
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    (Original post by OpenArms)
    Thanks for replying, is there any chance you could explain a bit further?
    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\left(x-\frac{\pi}{3}\right).

    Let f(x)=\sin x then 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 2\sin\left(x-\frac{\pi}{3}\right) = 2f(x-\frac{\pi}{3})
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    (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\left(x-\frac{\pi}{3}\right).

    Let f(x)=\sin x then 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 -?
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    (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 .

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

    See the edit in my previous post to see what to do next.
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    (Original post by OpenArms)
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    You're correct in how you've drawn \sin(x-\frac{\pi}{3}).

    Where does \sin x cross the x-axis? For every point where it crosses, \sin(x-\frac{\pi}{3}) will cross \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.
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    (Original post by notnek)
    You're correct in how you've drawn \sin(x-\frac{\pi}{3}).

    Where does \sin x cross the x-axis? For every point where it crosses, \sin(x-\frac{\pi}{3}) will cross \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  \frac\pi{3} ?

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    (Original post by OpenArms)
    So the graph is stretched to double upwards and has shifted to the right by  \frac\pi{3} ?

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

    The first transformation moves the maximum  \frac{\pi}{3} units to the right so it becomes (\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 (\frac{5\pi}{6},2).
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    (Original post by notnek)
    I'm not sure why the maximum of your graph has become 5. Did you draw \sin x correctly? The graph of \sin x is between -1 and 1 with the first maximum (starting from 0 and going right) at (\frac{\pi}{2},1).

    The first transformation moves the maximum  \frac{\pi}{3} units to the right so it becomes (\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 (\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.

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    (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.

    Click image for larger version. 

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    The maximum becomes 5 if you draw 2\sin(x-\frac{\pi}{3})+3 (which is where all this working is leading to) but I thought the graph you drew was of 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=0 into 2\sin(x-\frac{\pi}{3})+3.
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    (Original post by notnek)
    The maximum becomes 5 if you draw 2\sin(x-\frac{\pi}{3})+3 (which is where all this working is leading to) but I thought the graph you drew was of 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=0 into 2\sin(x-\frac{\pi}{3})+3.
    Apologies for the confusion.

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

    Here is the graph of 2\sin(x-\frac{\pi}{3})
    Click image for larger version. 

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    and here is the graph of 2\sin(x-\frac{\pi}{3})+3
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    plus some scribbling's on how to get the y value.
    Since sin is an odd function, sin(-\frac{\pi}{3}) = -sin(\frac{\pi}{3}).

    \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)?
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    (Original post by notnek)
    Since sin is an odd function, sin(-\frac{\pi}{3}) = -sin(\frac{\pi}{3}).

    \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)?
    Well I do the scottish system but I do know my exact values

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

    \sin(60) = \frac{\sqrt{3}}{2}
    Sorry, I'm getting my threads mixed up.

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

    So have you completed the question now?

    2sin(-\frac{\pi}{3}) + 3

    = -2sin(\frac{\pi}{3})+ 3

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

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

    = -\sqrt{3} + 3

    =  3 - \sqrt{3}

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

    = -2sin(\frac{\pi}{3})+ 3

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

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

    = -\sqrt{3} + 3

    =  3 - \sqrt{3}

    Is this correct?
    Yes.

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