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 confused ((Unless you notice the occasional error flagged up in the footnotes, obviously)). 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 $\cos(x) \ne 0$.