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I've managed to use identities to simplify it down to: $$r = 2\left(\cos^3\left({\theta\over2}\right)-3\sin\left({\theta\over2}\right)\cos\left({\theta\over2}\right)\right)$$ using trig identities, but I am stuck now. Does this have a Cartesian equivalent?

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Hint. The difficulty is the half angle. So: $$\eqalign{r^2 &=4\cos^2\Bigl(\frac{3\theta}{2}\Bigr)\cr &=2\bigl(1+\cos(3\theta)\bigr)\ .\cr}$$ Now you should be able to write in various ways $\cos(3\theta)$ in terms of $\cos\theta$ or $\sin\theta$ or both, and the rest should not be too difficult. Good luck!

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  • $\begingroup$ Of course! Why didn't I think of squaring the regular half angle identity. Thank you! I'll return when I've finished or run into another issue. $\endgroup$ – Keanu Vestil Feb 6 '14 at 5:06
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I want to elaborate on David's answer but please give him the credit...

From David's result $$ r^2 = 2 ( 4 \cos^3(\theta)- 3 \cos(\theta)+1)$$ Now $\cos(\theta) = x/r$ and hence $$ r^2 = {{8\,x^3}\over{r^3}}-{{6\,x}\over{r}}+2$$ cross multiplying we get $$-8\,x^3+6\,r^2\,x+r^5-2\,r^3=0$$ Now substitute $r =\sqrt{y^2+x^2}$ to get $$y^4\,\sqrt{y^2+x^2}+2\,x^2\,y^2\,\sqrt{y^2+x^2}-2\,y^2\,\sqrt{y^2+x ^2}+x^4\,\sqrt{y^2+x^2}-2\,x^2\,\sqrt{y^2+x^2}+6\,x\,y^2-2\,x^3=0$$

Here is the plot (remember to use $\theta = 0 \cdots 4 \pi$)

Plot

The above graph was produced using Octave (a free software similar to Matlab) Here are the commands:

theta=0:0.001:4*pi;
r=2*cos(3*theta/2);
plot(r.*cos(theta), r.*sin(theta))
print -djpeg keanu.jpg

Added more info in response to OP's question:

You can isolate $\sqrt{y^2+x^2}$, square it to get

$$-y^{10}-5\,x^2\,y^8+4\,y^8-10\,x^4\,y^6+16\,x^2\,y^6-4\,y^6-10\,x^6 \,y^4\\+24\,x^4\,y^4+24\,x^2\,y^4-5\,x^8\,y^2+16\,x^6\,y^2-36\,x^4\,y^ 2-x^{10}+4\,x^8=0$$ [I used maxima to do the work for me!]

You can sweep $x$ from $-2$ to $2$ and solve for $y$ using "roots" command and plot the result to get the pattern that you see.

This method is not very efficient but can be used for quick checks: Here is octave script to check the plot. It has some extraneous solutions that have to be pruned (complex roots) but that is a lot more work!

h=@(x)sqrt(roots([-1,4-5*x^2,-10*x^4+16*x^2-4,...
 -10*x^6+24*x^4+24*x^2,...
 -5*x^8+16*x^6-36*x^4,4*x^8-x^10]));
hold on;
for x=-2:0.005:2 plot(x*ones(5,1), h(x), '.'); end
for x=-2:0.005:2 plot(x*ones(5,1), -h(x), '.'); end
hold off
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  • $\begingroup$ Thank you for going through and solving this out. I came to the same result. $\endgroup$ – Keanu Vestil Feb 6 '14 at 5:34
  • $\begingroup$ When I graph this relation, should it resemble the original polar equation? $\endgroup$ – Keanu Vestil Feb 6 '14 at 5:45
  • $\begingroup$ yes, except you have an implicit function to graph. $\endgroup$ – user44197 Feb 6 '14 at 5:46
  • $\begingroup$ When I graphed it using CountourPlot[] in Mathematica, it only appeared as one half of the original graph. However, replacing all the x's with (-x)'s reflected it across the y axis, therefore completing it. Is this just an issue with the square roots present? $\endgroup$ – Keanu Vestil Feb 6 '14 at 5:51
  • $\begingroup$ Perhaps, you went for $\theta=0 \cdot \cdot 2 \pi$. You need to go to $4 \pi$ since you have half angles. See my output $\endgroup$ – user44197 Feb 6 '14 at 5:52

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