Angle of Intersection for two curves? Find the angle of intersection of ($r = 4 + \cos(3 \theta)$) and ($r = 4 -\cos(3\theta)$) at any point of intersection
Can someone provide me a hint to express the curve as a parameterized curve in Cartesian and treat $\theta$ as a the parameter variable?
 A: If I'm in a plane point $[y, x]$ where $y=f_1(x)$, then this point I can write $[f_1(x),x]$. [$[f_1^{'}(x), x]$ reprezents the tangent point in $x$. If $y=f_2(x)$ then $[f_2^{'}(x), x]$ reprezents also the tangent point in $x$(now for $f_2(x)$. If $x = a$ then points are $[f_1^{'}(a), a]$ $[f_2^{'}(a), a]$.
Is that $f^{'}(a) = \tan(\phi) = {{b}\over{c}} $. If $c=1$, $b=f^{'}(a)$, and it represents a vector $(1, f^{'}(a))$ with an angle $\tan(\phi)\equiv{f^{'}(a)}$, so $\phi = \arctan(f^{'}(a))$.  If a is intersect of $f_1$ with $f_2$ then angle of intersection is $$\arctan(f_1^{'}(a)) - \arctan(f_2^{'}(a)) $$
If $f_1 = 4 + \cos(3\theta)$ and $f_2 = 4 - \cos(3\theta)$
Now we find $a$ using equivalce: $4 + \cos(3\theta) = 4 - \cos(3\theta) \Rightarrow 2\cdot{cos(3\theta)} = 0 \Rightarrow cos(3\theta) = 0 \Rightarrow {\theta}=k{{\pi}\over{6}}$, where k=1, 2,...
$f_1^{'} = -3\sin(3\theta), f_2^{'} = 3\sin(3\theta)$ and $f_1^{'}(\theta) = -3\sin(k{\pi\over{2}}) = -3, f_2^{'}(\theta) = 3\sin(k{\pi\over{2}}) = 3$, these angles are the same only oposite. So searched angle is about $$143^{o}$$
A: You have the two curves
$$\eqalign{\gamma_1:\quad &\theta\mapsto {\bf f}(\theta):=\bigl(4+\cos(3\theta)\bigr)(\cos\theta,\sin\theta) \cr
\gamma_2:\quad &\theta\mapsto{\bf g}(\theta):= \bigl(4-\cos(3\theta)\bigr)(\cos\theta,\sin\theta)\ ,\cr}$$
where $0\leq\theta\leq 2\pi$, say.
As $4\pm \cos(3\theta)>0$  we have $\arg {\bf f}(\theta)=\arg{\bf g}(\theta)=\theta$ (modulo $2\pi$) for all $\theta$. Therefore the only way that $\gamma_1$ and $\gamma_2$ can intersect is when at a common parameter value $\theta$ we have $4+\cos(3\theta)=4-\cos(3\theta)$, or $\cos(3\theta)=0$. This is the case for the six $\theta$-values
$$\theta_k:={\pi\over 6}+k{\pi\over3}\qquad(0\leq k\leq 5)\ .$$
To find the angle of intersection at the point ${\bf z}_k:={\bf f}(\theta_k)={\bf g}(\theta_k)$ we have to compute the tangent vectors
${\bf t}_1:={\bf f'}(\theta_k)$, ${\bf t}_2:={\bf g}'(\theta_k)$. These two vectors are attached at ${\bf z}_k$ and point in the respective forward directions of $\gamma_1$ and $\gamma_2$ at this point. The angle between them can be found via the scalar product:
$$\angle({\bf t}_1,{\bf t}_2)=\arccos{{\bf t}_1\cdot {\bf t}_2\over|{\bf t}_1|\ |{\bf t}_2|}\ .$$
I leave the computation to the OP.
A: Well, both curves have their own set of points, $(r(\theta),\theta)$ and $(r(\theta),\theta)$. Let's set there $r$s equal.  
$$4+\cos(3\theta)=4-\cos(3\theta)$$
Then we have $2\cos(3\theta)=0$. Then $3\theta=\dots,{\pi\over2},{3\pi\over2},{5\pi\over 2},{7\pi\over2},\dots$. So $\theta={\pi\over6},{\pi\over2},{5\pi\over6},{7\pi\over6},{3\pi\over2},{11\pi\over6}$. 
BUT THAT WASN'T YOUR QUESTION:::
Here's a hint, how about $(x,y)=(r\cos(\theta),r\sin(\theta))$? 
