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Polar to Rectangular Conversion

  1. May 1, 2005 #1
    Can someone help me with these problems? It's been bugging me i can't seem to solve it.

    Lets assume T = theta

    I can't seem to find a way to convert these polar equations into rectangular form.

    r = 2 sin 3T

    r = 6 / 2 - 3 sinT

    If possible, can someone help me with this and list it in a step-by-step fashion so i can see how you get the answer. Thanks.
    Last edited: May 1, 2005
  2. jcsd
  3. May 1, 2005 #2


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    [tex] r = 2 \sin (3 \theta) [/tex]

    [tex] r^2 = 2r \sin (3 \theta) [/tex]

    then try the triple angle formulas to reduce [itex] \sin (3 \theta) [/itex]
  4. May 1, 2005 #3
    Sorry but the furthest that our teacher had taught us in the double angle formula.
  5. May 1, 2005 #4


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    It's not hard. Why don't you try it?

    [tex] \sin (3 \theta) = \sin (\theta + 2 \theta) [/tex]

    Use the sum of angles for the sin

    [tex] \sin (\alpha + \beta) = \sin \alpha cos \beta + \cos \alpha \sin \beta [/tex]
  6. May 1, 2005 #5
    Ok, so i got:

    [tex] r^2 = 2r (\sin \theta \cos 2 \theta + \cos \theta \sin 2 \theta ) [/tex]

    can i take out the [tex] \sin \theta \cos \theta [/tex]

    and get

    [tex] r^2 = 2r \sin \theta \cos \theta (\cos \theta + \sin \theta ) [/tex]

    or should i further expand the equation?
  7. May 1, 2005 #6


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    No that's wrong it will be

    [tex] \sin (3 \theta) = 3 \sin \theta - 4 \sin^{3} \theta [/tex]


    [tex] r^2 = 2r (3 \sin \theta - 4 \sin^{3} \theta) [/tex]
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