A trigonometry masterclass

I mean, I don’t mean to imply that I’m a master, although I’m pretty good at A-level trig. I just wanted to talk through my thought process in solving a question, and “masterclass” seemed like a good word.

The question is:

(a) Show that $\frac{1}{\cos (\theta )}+\tan (\theta )\equiv \frac{\cos (\theta )}{1-\sin (\theta )},\theta \ne (2n+1){90}^{\circ },n\in \mathbb{Z}$

Given that $\cos (2x)\ne 0$

(b) Solve for $0<x<{90}^{\circ }$:

$\frac{1}{\cos (2x)}+\tan (2x)=3\cos (2x)$

It’s three marks for (a) and five for (b), which I don’t really care about, but tends to give an idea of how much work is expected.

OK! Let’s start with the show that bit.

My first thought is, $\tan (\theta )$ is the same as $\frac{\sin (\theta )}{\cos (\theta )}$, which makes the left-hand side quite simple – but an annoyingly different form to the right-hand side. I’ll work it out: the left-hand side is $\frac{1}{\cos (\theta )}+\frac{\sin (\theta )}{\cos (\theta )}$, or $\frac{1+\sin (\theta )}{\cos (\theta )}$.

That’s not (immediately) the same as $\frac{\cos (\theta )}{1-\sin (\theta )}$, though. I could check, if I wanted – put in a random value of $\theta$ on both sides and see if I get the same. If I’ve made an error, this will most likely tell me about it. It doesn’t prove anything, though.

My spidey senses tingle at seeing both a $1-\sin (\theta )$ and a $1+\sin (\theta )$ – multiplying those would give me a ${\cos }^2(\theta )$, and a little bit of mental cross-multiplication convinces me that that’s exactly what we’d get. Unfo rtunately, that’s not a permitted move in a show that question.

What is permitted is to multiply top and bottom by the same, non-zero thing: so I can do something like $\frac{1+\sin (\theta )}{\cos \theta }\cdot \frac{1-\sin (\theta )}{1-\sin (\theta )}$ – so long as $\sin (\theta )\ne 1$. Fortunately, that ‘s what the condition after the equivalence means, so it’s legit. Multiplying it out gives $\frac{(1+\sin (\theta ))(1-\sin (\theta )}{\cos (\theta )(1-\sin (\theta ))}$, or $\frac{1-{\sin }^2(\theta )}{\cos (\theta )(1-\sin (\theta )}$.

We’re nearly there! If we replace the top with ${\cos }^2(\theta )$ and cancel a $\cos (\theta )$– which we can do, since the condition also implies that $\cos (\theta )\ne 0$ – we get $\frac{\cos(\theta)}{1-\sin(\theta)$. Boom!

The second part does something a lot of questions like this do: reuse the result from the first part in the second. (It’s worth noting that you can use the result even if you didn’t get part (a) – you don’t have to miss out on these marks just because you didn’t get the first bit right.) It’s the same sort of shape – you just need to play spot the difference and notice that $\theta$ has become $2x$.

I’d prefer to use $\theta$ – and since $0<x<{90}^{\circ }$, we can say $0<\theta <{180}^{\circ }$ (since it’s the same as $2x$.

Now rewrite it as $\frac{1}{\cos(\theta) + \tan(\theta) = 3 \cos(\theta)$andreplacetheLHSwiththeresultfromearlier:$\frac{\cos(\theta)}{1-\sin(\theta)} = 3 \cos(\theta)$.

We’re told that $\cos (\theta )\ne 0$, so we can divide it out: $\frac{1}{1-\sin (\theta )}=3$, or $1-\sin (\theta )=\frac{1}{3}$. Better yet, $\sin (\theta )=\frac{2}{3}$, and the calculating machine gives an answer of ${41.8}^{\circ }$ for $\theta$. ¹

It would be easy – and wrong – to write down 41.8 degrees and think “job’s a good-un”.

It would be easy – and incomplete – to notice that we made $\theta$ up and we need to halve it to get $x$, and write down ${20.9}^{\circ }$.

The correct thing to do is to draw a small graph of $y=\sin (\theta )$ and notice that there’s a second solution at $\theta =(180-41.8{)}^{\circ }$. That works out to $x=(90-20.9{)}^{\circ }$, or ${69.1}^{\circ }$.

There are two solutions: 20.9 and 69.1 degrees, and it’s good practice to put them (or better, the exact values) back into the equation to check.

Would you have done it differently? Was there anything I missed? Do let me know!

1. I pretend it was the calculating machine. It’s actually a value I have memorised in case the Ninja is around. ↩