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Sin curve

  1. Nov 3, 2005 #1
    is there a way to turn a sin curve (sin(x)) from a horizontal graph to a 45 degree, y=x graph mathematically
     
  2. jcsd
  3. Nov 3, 2005 #2

    James R

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    Yes.

    Your initial equation is y=sin x

    Put

    [tex]x = x' \cos \theta - y' \sin \theta[/tex]
    [tex]y = x' \sin \theta + y' \cos \theta[/tex]

    where, in your case [itex]\theta = 45[/itex] degrees.

    Then, your new equation is:

    [tex]\frac{1}{\sqrt{2}}(x' + y') = \sin \left[ \frac{1}{\sqrt{2}}(x' - y') \right][/tex]

    Now, the only problem you have is to rearrange this so as to write y' in terms of x'.
     
  4. Nov 4, 2005 #3

    TD

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    You rotated over an angle of -45° because your minus-sign was placed wrongly. You have to be careful whether it's the coordinate axis which are being rotated or the function itself. In this case, the rotation matrix is:

    [tex]\left( {\begin{array}{*{20}c}
    {\cos \theta } & {\sin \theta } \\
    { - \sin \theta } & {\cos \theta } \\
    \end{array}} \right)[/tex]

    giving the new equation:

    [tex]\frac{1}{\sqrt{2}}(y' - x') = \sin \left[ \frac{1}{\sqrt{2}}(x' + y') \right][/tex]
     
  5. Nov 6, 2005 #4

    James R

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    Fair enough.
     
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