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Deriving equation for circle using sin and cos identities

  1. Sep 17, 2008 #1
    1. The problem statement, all variables and given/known data
    I have been given these two equations:
    x=2acos^2(x) , y = 2a(cos(x))(sin(x)) where a ranges from 0 to 5 and -2π < x < 2π
    I need to prove that these equations (when you plug in values for x) create points that when plotted, give you a circle with center (x-a) and radius a.

    2. Relevant equations
    I have gotten the correct answer, however I am missing a step and I'm not sure what I did to get the correct answer.

    3. The attempt at a solution
    I wanted to get an equation for a circle out of the two equations I have been given. So using the sin and cos identities, I can get rid of the x's. I know 2cos^2x = cos2x + 1 and that 2(cos(x))(sin(x)) = sin2x. I then added the a values back in so I have these two equations:
    x= a+a(cos(2x)) , y= a(sin(2x))
    From there, I fixed the x value to look like this:
    (x-a) = a(cos(2x)) , y = a(sin(2x))
    Then I remembered the circle equation of x^2 + y^2 = 1
    So I did this:
    (x-a)^2 + y^2 = a^2(cos^2(2x)) + a^2(sin^2(2x))
    I know I'm supposed to get (x-a)^2 + y^2 = a^2 , but I'm not sure what to do to the above equation to get this result. Any help would be greatly appreciated!
  2. jcsd
  3. Sep 17, 2008 #2
    Try factoring out a^2 from the right hand side.
  4. Sep 17, 2008 #3


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    What do you know about

    [tex] \sin^2 \theta + \cos^2 \theta [/tex] ?
  5. Sep 17, 2008 #4
    Oh, I get it. So I could write it like this:

    (x-a)^2 + y^2 = a^2 (cos^2(2x) + sin^2(2x))

    Then the identity takes care of the rest so I'm left with

    (x-a)^2 + y^2 = ^2 , right?
  6. Sep 17, 2008 #5
    I meant to write the above post as:

    (x-a)^2 + y^2 = a^2
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