I was coaching a student on how to approach this problem. 2 equations given and the question was where they intersect.

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Now, with a bit of practice on polar coordinates, producing the graph by hand wasn't too tough. I was able to point out the petal in the first quadrant wan't going to intersect the vertical line unless rotated 30 degrees clockwise.

The student said the result was expected to be algebraic, i.e. set the 2 equal, manipulate and show no solution. I wasn't able to do this after about 10 minutes of manipulation, I just would up with sine to the 3rd power along with a number of sine*cos terms.

I'm not so much looking for this single solution as I am a process how to approach this style of problem. (This is for a pre-calc class, typically with high school juniors)


4 Answers 4


This might not quite be an "algebraic" solution in the sense the student was looking for, but I think a more reasonable approach to this sort of problem is using bounds, not setting things equal from the outset.

The general strategy is to notice that one curve is entirely inside a circle of radius 4, while the other is entirely outside that circle, so any intersection must occur on the boundary. This is often a good strategy for showing two objects don't intersect: bound them inside larger, simpler regions that can be shown to intersect only in certain very limited ways.

More precisely, since $\sec(\theta) = 1/\cos(\theta)$, we have $$\lvert 4\sec(\theta) \rvert \geq 4$$ with equality if and only if $\lvert \cos(\theta) \rvert = 1$. However, $$\lvert 4 \sin(3 \theta) \rvert \leq 4,$$ so if $4 \sec(\theta) = 4 \sin(3\theta)$, then $\lvert \cos(\theta) \rvert = 1$, which means $\theta$ is an integer multiple of $\pi$. But that implies $4 \sin(3\theta) = 0$, a contradiction.

  • $\begingroup$ Yes, not as algebraic as say "Sin X = 2 therefore no solution," but a pretty elegant answer in that it touches on what they should know about setting up proofs. Thx. $\endgroup$ Commented Dec 7, 2016 at 14:12

What about taking what @whatever got: $ (4\cos^2(\theta)-1)^2(1-\cos^2(\theta)) \cos^2(\theta)=1$.

Let $ x = \cos(\theta).$ This gives us a more familiar looking: $(4x^2-1)^2(1-x^2)x^2=1$, which we can expand (and clear leading negatives for appearances sake):

$16 x^8 - 24 x^6 + 9 x^4 - x^2 + 1=0$

Now, if your students are capable of doing polar coordinates at the level you're doing, then most assuredly they've been exposed to dealing with polynomials and their roots.

A simple application of the Rational Root Test shows no rational roots.

Then, applying Descartes's Rule of Signs says:

4 possible real roots, and hence 4 possible complex solutions.

It's an even-degree polynomial with positive leading coefficient. Without techniques from calculus or non-familiar things, however, they should do exactly what all of us would: graph that polynomial on their calculators. No zeroes. Therefore the original question is answered, because a polynomial they're familiar with has no (real) roots.

This could be a place to allude to the fact that even though the polar graphs don't intersect in the real plane, they have DO have an intersection -- in the complex plane, just like $y=-1$ and $y=x^2$ intersect in the complex plane but not the real one.


$R=4\sec(\theta)$ is the same as $x=4$. The polar curve $R=4\sin(3\theta)$ is contained in the circle $R=4$. The only possible intersections of the vertical line and the circle occur when $3\theta=\pm\pi/2,\pm3\pi/2,\cdots$. So the polar curve and the vertical line do not intersect.


$$\sin(3\theta)=\sec \theta$$ $$\sec \theta=(4\cos^2 \theta-1)\sin \theta$$ $$(4\cos^2 \theta-1)\sin \theta \cos \theta=1$$ $$(4\cos^2 \theta-1)^2(1-\cos^2 \theta) \cos^2 \theta=1$$ which is an eighth degree polynomial in $\cos \theta$.

  • 1
    $\begingroup$ I got a similar mess, now, how to prove no solution? $\endgroup$ Commented Dec 2, 2016 at 17:43
  • $\begingroup$ just show that the global maximum of the polynomial function is negative. $\endgroup$
    – user5402
    Commented Dec 2, 2016 at 18:10

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