Written by RealityMinus3+ in dialogue, trigonometry.

“You know how you’re always putting things like ‘just to keep @RealityMinus3 happy’ in your posts?”

“Of course, sensei!”

“Well… you remember that post about missing solutions in a trig problem?”

“Ut-oh.”

*What follows is a guest post by Elizabeth A. Williams, who is @RealityMinus3 in real life.*

This thing about “infinite denominators”: I solidly feel that this is not the way to think about it, and it’s not what I would encourage a new-material student to do, because “infinite denominators” are an effect (usually, perhaps always?) of dividing by zero, and the thing to do is to *think about not dividing by zero*.

Infinity is a slippery bastard, full of contradictions. Even an army of staple guns has trouble pinning that biter down. It’s not even just one biter! Night. Mare. Also, it’s good (nay, necessary) practice in general to think about what a function will and will not allow; that, in a direct approach, should always be in the forefront of a mathematician’s mind.

When I sit down with an equation — in its initial state — one of the first things I consider is whether there are values that the variable cannot take. For example:

- even-root-taking of a negative quantity;
- arguments of functions invalid (eg logs);
- division by zero;
- etc.

First off, I note that $0 \le x \le 2\pi$ which is a “because I say so” restriction rather than “mathematics says no” – an Important Distinction, I think.

This equation, $2\tan(2x) – 2\cot(x)=0$, has tangent and cotangent in it, so I would list the values that are disowned by these functions. I reiterate that I am going by the original equation — which was my key^{1} to losing and then recovering those missing solutions.

Right, so,

The $\tan(2x)$ means $x$ cannot be $\piby 4$, $\frac{3}{4}\pi$, $\frac{5}{4}\pi$, or $\frac{7}{4}\pi$.

Also, $\cot(x)$ means $x$ cannot be $0$, $\pi$, or $2\pi$.

Now to trig algebra this to its grim death… I mean, and now to solve the equation.

Going via the double angle formula, $\tan(2x)$ converts into an expression involving $\tan(x)$. **It is at this point that we lose the possibility of $x$ being $\piby 2$ or $\frac{3}{2}\pi$.** Bang!

Bang bang! Stop to make a note about that, right here in the algebra, to go back and check these excluded values in the original equation, just to make sure.

If we stay in tangent, then $\piby 2$ and $\frac{3}{2}\pi$ will not re-appear as possibilities. Onwards, and we get

$x = \piby 6, \frac{5}{6}\pi, \frac{7}{6}\pi$, and $\frac{11}{6}\pi$.

(For interest, I said:

$\frac{2 \tan (x)}{ 1 – \tan^2 (x)} = \cot(x)$

Then, because $\cot(x) = \frac{1}{ \tan (x)}$, $2 \tan^2 (x) = 1 – \tan^2 (x)$

$3 \tan^2 (x) = 1$

$\tan (x) = \pm\frac{\sqrt{3}}{ 3}$)

(I note, out of an abundance of noting, that $1 – \tan^2 (x)$ in the denominator of the double-angle tangent formula excludes $x$ from being $\piby 4$ and so on in the very same way that $\tan(2x)$’s $\cos(2x)$ does.)

Check the missing $\piby 2$ and $\frac{3}{2}\pi$ — they work — and viola (you heard me), six solutions.

If you go from plain tangents into sine and cosine though … Well. Colin. I cannot believe you allowed cosine to be a denominator without noting that this excludes $x$ from being $\piby 2$ or $\frac{3}{2}\pi$. [Yes miss sorry miss – ed.] Similarly, rewriting cotangent means sine in the denominator, which leads to more excluded values.

In both cases, this should have already been picked up right at the start because tangent and cotangent domains. It’s just more obviously explicit (YOU HEARD ME) now.

When one multiplies anything through by cosine-squared, whether it came from a denominator or not, one ought to be careful right here about whether one might be multiplying both sides by zero. Eek. In fact, I think you did implicitly multiply by zero, because cosine ends up in the numerator from this, where it ostensibly causes no domain issues at all. In effect, you’ve gone from tangent to cosine-on-top, and this forgives tangent’s restrictions.

It’s really interesting that $\tan(2x)$ to $\tan(x)$ lost bits, but then bumping up by cosine got them back again – but that might be another exploration beyond this post.

(As an aside: note that the issue of multiplying both sides of a relation by a function rather than a hard number, or taking a function of both sides, or, or, or … is by no means simple, straightforward, novice-ranked territory.)

It’s excellent that you factor out a cosine instead of dividing both sides of the equation by cosine. That is good. You would lose the solutions again at this point if you did that. I am amused though! You’re careful to check solutions to the equation for cosine equaling zero; but it’s because of not being careful with zero earlier that got you here.

I am *horrified* at “[fraction-expression = zero] This is only true when either the bottom goes to infinity (which it never does)…”. Aaaaaaaagh. No. Please. Fractions equal zero only when their numerators equal zero. If the limit of their denominators is unbounded as the variable goes to some number, the fraction as a whole *might* tend towards zero … but asymptotically. Mind the epsilon!

Also, it’s not as $x$ approaches $\piby 2$ or $ \frac{3}{2}\pi$ that the fraction is undefined: the fraction is very well defined for any of these values *on the approach*. It’s when $x$ *equals* these values that the problem arises.

But your key is right there in “meaning those values also warrant further investigation”. Bang bang!

I’m not sure about your “Or does it?” section. $x$ being $\piby 2$ works perfectly, and the emphasis here is “use the original equation”. It doesn’t sit poorly at all. I’m not sure about rewriting everything in cotangent being a fix for feeling better about the original, especially as $\cot \br{\piby 2}$ equals 1, so that denominator there would be zero. [Good point, well made – ed.]

- I want to say “the key” because I think this approach is the clean and rigorous way of looking at it, but I’m not sure about shouldering an announced exclusive on that without agreement from elsewhere. [↩]