Some interesting confusion

I have a tendency to write about interesting questions from a ‘here’s how you do it’ point of view, which must give the impression that I never get confused1. To try to dispel that, I wanted to share something that came up in an Oxford entrance paper (the MAT from 2010, if you’re interested).

It’s not that I got the question wrong immediately, more that my student took a different approach, got a different answer, and it took us a while to figure out why.

The question was to find how many solutions to $\sin^2(x) + 3 \sin(x)\cos(x) + 2 \cos^2(x)=0$ there are for $0 \le x \lt 2\pi$.

(In the spirit of here’s how to do it: it factorises nicely as $\left(\sin(x) + \cos(x)\right)\left(\sin(x) + 2\cos(x)\right)=0$, which gives two possibilities: $\tan(x) = -1$ or $\tan(x) =-2$, and there are two possibilities for each of those making four altogether.)

Instead, my student did something quite interesting: he said “I know $\sin(x) \equiv \tan(x) \cos(x)$, so I can rewrite this as $\tan^2(x) \cos^2(x) + 3 \tan(x) \cos^2(x) + 2 \cos^2(x) = 0$. Then I can factor out the $\cos^2(x)$ to leave $\cos^2(x) \left( t^2 + 3t + 2 \right) = 0$, where $t = \tan(x)$.”

As with my method, the quadratic bracket factorises to give $t = -1$ or $t = -2$, and four solutions there; however, $\cos^2(x) = 0$ twice as well! That makes six altogether.

What gives?

I encourage you to spend a few moments thinking about it before reading on, unless it’s obvious to you, in which case yah-boo, nobody likes a smart-arse.

The problem was with the identity: it’s not universally true that $\sin(x) \equiv \tan(x) \cos(x)$. It breaks down periodically — specifically, it only works where $\tan(x)$ is defined. Where is $\tan(x)$ undefined? Precisely where $\cos(x) = 0$, because $\tan(x) \equiv \frac{\sin(x)}{\cos(x)}$ and you can’t divide by zero.

In this particular case, when $\cos(x) = 0$, the quadratic bracket is undefined. We can get around this by seeing what happens when $\cos(x) = 0$ goes into the original equation: we get $\sin^2(x) = 0$, which isn’t true if $\cos(x) = 0$.

So, what’s the moral here? It’s to be careful of the conditions on your identities. The expressions $\sin^2(x) + 3 \sin(x)\cos(x) + 2 \cos^2(x)$ and $\cos^2(x) \left( \tan^2(x) + 3\tan(x) + 2 \right)$ are equivalent only when $\cos(x) \ne 0$.


Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

  1. Unless you notice the occasional error flagged up in the footnotes, obviously []


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