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Math Help - convert polar equation to rectangular form: r=2sin3θ

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    convert polar equation to rectangular form: r=2sin3θ

    i have no idea on how to go about solving this, please help!
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    Quote Originally Posted by got_jane View Post
    i have no idea on how to go about solving this, please help!
    Let's call (theta) = t for convenience.

    sin(3t) = sin(2t + t) = sin(2t)cos(t) + sin(t)cos(2t)

    = 2sin(t)cos(t)*cos(t) + sin(t)*(1 - 2sin^2(t))

    = 2sin(t)cos^2(t) + sin(t) - 2sin^3(t)

    = 2sin(t)(1 - sin^2(t)) + sin(t) - 2sin^3(t)

    = 2sin(t) - 2sin^3(t) + sin(t) - 2sin^3(t)

    = 3sin(t) - 4sin^3(t)

    So your polar equation is:

    r = 6sin(t) - 8sin^3(t)

    Now,
    r = (x^2 + y^2)^{1/2}
    t = atn(y/x)

    This is probably not the easiest way to go. I mean, sin(atn(y/x))? It can be done, but we have to do this for four different quadrants. Let's try this:
    y = r*sin(t)

    sin(t) = y/r = y/(x^2 + y^2)^{1/2}

    So...
    (x^2 + y^2)^{1/2} = 6*y/(x^2 + y^2)^{1/2} - 8*y^3/(x^2 + y^2)^{3/2}

    I'm going to clear the fractions and multiply both sides by (x^2 + y^2)^{3/2}:
    (x^2 + y^2)^2 = 6y(x^2 + y^2) - 8y^3

    And you can simplify from here as you like. It's not going to simplify into anything pretty anyway.

    -Dan
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