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Math Help - Applied Algebra x,y coordinates please help

  1. #1
    Member cmf0106's Avatar
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    Applied Algebra x,y coordinates please help

    It reads: When you rotate the x-y coordinate system by an angle of \theta, the new cordinates x' and y' can be calculated using the following expressions:

    x'= xCos \theta+ ysin \theta (1)
    y'=-xsin \theta + ycos \theta(2)

    Given that you know the rotated coordinates x' and y' and the rotation angle \theta how do you calculate the original x an y? (in other words, how do you express x and y using x', y' and \theta?

    Hint: rearrange equations (1) and (2) so that x and y can be expressed using x', y', sin \theta and cos \theta

    and thats all it gives me except for the addition of another example problem which asks

    if x'=24, y'=-3 and \theta=30 degrees, what are the values of x and y?

    many thanks, also what are these problems trying to show im not really sure what they have to do with college algebra.
    Last edited by cmf0106; September 21st 2008 at 07:07 PM.
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  2. #2
    MHF Contributor
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    It is just reversing the procedure.

    To get the (x',y') from (x,y) with a rotation of theta, the following were derived:
    x'= xCos(t) + ysin(t) --------(1)
    y'= -xsin(t) + ycos(t) ..........(2)

    So if you reverse the procedure, to get back to the (x,y), you use
    x = x'Cos(-t) + y'sin(-t) --------(3)
    y'= -x'sin(-t) + y'cos(-t) ..........(4)
    where the angle of rotation is the reverse of the theta before.

    Rewriting those, in terms of the original theta,
    x = x'cos(t) -y'sint(t) -------(3a)
    y = x'sin(t) +y'cos(t) --------(4a)

    that is because cos(-t) = cos(t), and sin(-t) = -sin(t).

    --------------------------
    '"if x'=24, y'=-3 and =30 degrees, what are the values of x and y?"

    x = 24cos(30deg) -(-3)sin(30deg) = 22.28461
    y = 24sin(30deg) +(-3)cos(30deg) = 9.40192

    Maybe you want to check if (22.28461,9.40192) will become (24,-3) if the original axes are rotated by 30 degrees.
    Last edited by ticbol; September 21st 2008 at 07:22 PM.
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  3. #3
    Member cmf0106's Avatar
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    Thank you sir! It is crystal clear now
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