Deceptive intersection of polar curves

Question

Consider the graphs of $r=\cos 2\theta$ and $r=\cos\theta -1$.

I would like to find all the intersections for the two graphs.

Equating both and solve I only manage to find $(-1,\frac{\pi}{2}),(-1,\frac{3\pi}{2}),(-\frac{1}{2},\frac{\pi}{3}),(-\frac{1}{2},\frac{5\pi}{3})$.

However, there seem to be other intersections as well as the two graphs may not meet at the same value of $\theta$.

Is there a way to solve for the other points of intersection?

Answer 1

In cartesian coordinates $(x,y)$, 2 points $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$ are equal if and only if there coordinates match exactly ($x_1=x_2, y_1=y_2$). Polar coordinates ($\theta,r$) are different, two points $P_1(\theta_1,r_1),P_2(\theta_2,r_2)$ can represent the same point in the plane without both components matching exactly.

The obvious case is $r_1=r_2, \theta_1-\theta_2=2k\pi, k\in \mathbb Z$. But you know that already and have (as usual) only consisdered $\theta$ in some interval of length $2\pi$, like the usual $[0,2\pi)$.

The first non-trivial case is that any point $(\theta,0)$ represents the origin of the coordinate system. Since $(\frac{\pi}4,0)$ is on the first curve $r=\cos2\theta$ and $(0,0)$ is in the second curve $r=\cos\theta-1$, both curves meet also at the origin.

The second non-trivial case is $\theta_1-\theta_2=\pi, r_1=-r_2$. If you allow negative $r$ (and your formulas will produce negative $r$ for both curves), changing the sign of $r$ means you reflect the point on the origin, resulting in $\theta$ changing by $\pi$. Another way to see that is if you convert back to cartesian coordinates:

$$ \begin{eqnarray} x_1 &=& r_1\cos\theta_1 &=&-r_1(-\cos\theta_1) &=& -r_1\cos(\theta_1-\pi) &=& r_2\cos\theta_2 &=& x_2\\ y_1 &=& r_1\sin\theta_1 &=&-r_1(-\sin\theta_1) &=& -r_1\sin(\theta_1-\pi) &=& r_2\sin\theta_2 &=& y_2\\ \end{eqnarray} $$

So, what needs to be done now?

1) You already solved one case, where $r_1=r_2, \theta_1-\theta_2=2k\pi, k\in \mathbb Z$.

2) The case $r_1=r_2=0$ I solved for you, it's usually very easy.

3) The case $\theta_1-\theta_2=\pi, r_1=-r_2$ is still open. It leads to the equation

$$\cos(2(\theta_2+\pi)) = r_1 = -r_2 =-(\cos\theta_2 - 1)$$

which I trust you can solve yourself, just like in case 1).

EDIT: I wrote that there is small error in your solution, but the error was on my part.

Answer 2

Hint: Use that $$\cos(2x)=2\cos^2(x)-1$$

Answer 3

$\cos 2\theta = \cos\theta - 1\\ 2\cos^2 \theta - 1 = \cos\theta - 1\\ \cos\theta(2\cos\theta - 1) = 0\\ \theta = \frac {\pi}{2},\frac {3\pi}{2}, \frac {\pi}{3}, \frac {5\pi}{3}$

However, the curves will also overlap when $r(\theta)$ in one curve equals $-r(\theta + \pi)$ in the other.

$\cos 2\theta = \cos\theta + 1\\ 2\cos^2 \theta - 1 = \cos\theta + 1\\ 2\cos^2 \theta - \cos\theta - 2$

$\cos \theta = \frac {1 - \sqrt {17}}{4}$

Which will give two more points.

The curves $r = \cos \theta - 1$ and $r=\cos\theta + 1$ are identical curves.

And the curves intersect at $r = 0$

Answer 4

Each non-zero polar coordinate $(r,\theta)$ has an equivalent form $(-r,\theta+\pi)$. Applying this substitution to the two equations yields the new equations $r=-\cos2\theta$ and $r=\cos\theta+1$. Writing $\cos\theta=s$, these become $$r=2s^2-1\tag1$$ $$r=1-2s^2\tag2$$ $$r=s-1\tag3$$ $$r=s+1\tag4$$ The four combinations $(\{1,2\},\{3,4\})$ of these equations need to be considered separately, and yield the following valid solutions $(r,s)$ where $s\in[-1,1]$: $$(1,3)\to (-1,0),\left(-\frac12,\frac12\right)$$ $$(1,4)\to\left(\frac{5-\sqrt{17}}4,\frac{1-\sqrt{17}}4\right)$$ $$(2,3)\to\left(\frac{\sqrt{17}-5}4,\frac{\sqrt{17}-1}4\right)$$ $$(2,4)\to(1,0),\left(\frac12,-\frac12\right)$$ Negating both numbers in a pair above yields the same point by the aformentioned equivalence of polar coordinates (for $\cos(\pi+\theta)=-\cos\theta$), so we have the following distinct pairs with $r>0$: $$(1,0),\left(\frac12,-\frac12\right),\left(\frac{5-\sqrt{17}}4,\frac{1-\sqrt{17}}4\right)$$ The first two pairs above correspond to the four non-origin solutions already found: $(r,\theta)=\left(1,\pm\frac\pi2\right),\left(\frac12,\pm\frac{2\pi}3\right)$. The last one corresponds to the missing two solutions $(r,\theta)=\left(\frac{5-\sqrt{17}}4,\pm\cos^{-1}\frac{1-\sqrt{17}}4\right)$. The last intersection of the two polar curves is the origin.