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Math Help - Divide 2Pi into 4

  1. #1
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    Dividing 2pi

    Hi, how can I divide 2pi into four equal parts?
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  2. #2
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    Please clarify

    I assume you do not simply mean \frac{2\pi}{4}
    Do you mean cutting an angle into four equal parts?

    2 pi radians implies a circle. You could bisect the circle and then bisect each portion once more.
    Last edited by Math Help; June 26th 2005 at 08:45 AM.
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  3. #3
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    I am trying to follow the logic of an example in my book. The example is Using Key Points to Sketch a Sine Curve.

    Sketch the graph of y = 2 sib x on interval [-pi, 4pi].

    it goes on to say: Note that y = 2 sub x = 2(sin x) indicates that the y-values for the key points will have twice the magnitude of the graph of y = sin x. Divide the preiod 2pi into four equal parts to get a set of key points. The points are: (0,0) , (pi/2,2) , (pi,0) , (3pi/-2) , and (2pi,0)

    How did the book come up with this set of points. Do I have to use a graphing calulator to find these points? Can I do this without a calculator?

    Thanks for any help with this one.
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  4. #4
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    Let us forget about the
    y = 2 sib x
    and
    y = 2 sub x

    They just add to the confusion.
    (What are those, anyway? )

    You mean
    y = 2 sin x
    for the key points?

    Well, that sure means the y-values for the key points are twice those corresponding y-values on the y = sin x.

    Umm, wait a minute,.....you mean
    y sub 2 = 2 sin x?
    Or,
    y_2 = 2 sin x?

    Because y = sin x?

    And so, y_2 = 2*y = 2 sin x.

    Figures.

    ---------------------
    And then you are asking how the book got the points
    (0,0), (pi/2,2), (pi,0), (3pi/2, -2) and (2pi,0)
    for y sub 2?

    -----------------
    In graphing y = sin x, usually the points at x = 0, pi/2, pi, 3pi/2 and 2pi are plotted on the x,y axes.

    (Like the 2pi was divided into 4 equal parts.
    2pi/4 = pi/2.
    So, 1st part is from 0 to pi/2
    2nd part is from pi/2 to pi
    3rd part is from pi to 3pi/2
    4th part is from 3pi/2 to 2pi.)

    Then,
    when x=0, y = sin(0) = 0. ...............so point (0,0)
    when x=pi/2, y = sin(pi/2) = 1. .......so point (pi/2,1)
    when x=pi, y = sin(pi) = 0. ..............so point (pi,0)
    when x=3pi/2, y = sin(3pi/2) = -1. ....so point (3pi/2,-1)
    when x=2pi, y = sin(2pi) = 0. ...........so point (2pi,0)

    ------
    Now, since y sub 2 = 2*y, then,
    at x=0, y sub 2 = 2*0 = 0 .............so point (0,0)
    at x=pi/2, y sub 2 = 2*1 = 2 ..........so point (pi/2,2)
    at x=pi, y sub 2 = 2*0 = 0 .............so point (pi,0)
    at x=3pi/2, y sub 2 = 2*-1 = -2 ......so point (3pi/2,-2)
    at x=2pi, y sub 2 = 2*0 = 0 ............so point (2pi,0)

    ------------
    That is how.
    Last edited by ticbol; June 27th 2005 at 05:55 AM.
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  5. #5
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    Thank you. That helps me so much. I appreciate all the information.
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