# Ask Uncle Colin: A Trig Identity Puzzle

Dear Uncle Colin,

Given that $\sin(x)\cos(x) = \frac{2}{5}$, how can I work out $\tan(x)$?

- Tangent Ratio, I Guess

Hi, TRIG, and thanks for your message!

I think this calls for some problem solving out loud! I see two immediate approaches.

### Approach 1: double angles

Whenever I see the product of sine and cosine, I jump straight to $\sin(2x)\equiv 2\sin(x)\cos(x)$ – in this case, we’d have $\sin(2x)=\frac{4}{5}$.

That looks like a 3-4-5 triangle, so $\cos(2x) = \pm \frac{3}{5}$ and $\tan(2x) = \pm \frac{4}{3}$.

We also know that $\tan(2x) \equiv \frac{2\tan(x)}{1-\tan^2(x)}$. Let’s let $t = \tan(x)$ and solve:

- $\pm \frac43 = \frac{2t}{1-t^2}$
- $4 - 4t^2 = \pm6t$
- $0 = 2t^2 \pm 3t - 2$
- $0 = (2t\mp1)(t\pm2)$

So, we have that $\tan(x)$ could be $\pm 2$ or $\pm \frac{1}{2}$. If we knew that we were looking in the first quadrant, we might be able to decide which of those worked (I’d imagine $\frac{1}{2}$).

### All the squares!

If $\sin(x)\cos(x) = \frac{2}{5}$ then $\sin^2(x)\cos^2(x) =\frac{4}{25}$.

Or rather, $25\left(1-\cos^2(x)\right)\cos^2(x)=4$. Which is ugly and confusing, so let’s let $\cos(x) = c$.

- $25(1-c^2)c^2 = 4$
- $0 = 25c^4 - 25x^2 + 4$
- $0 = (5c^2 - 1)(5c^2 - 4)$

So $\cos^2(x) = \frac{1}{5}$ or $\frac{4}{5}$ – and $\sec^2(x) = 5$ or $\frac{5}{4}$.

That’s useful because we can use Santa’s identity((“You even know the hard ones / you barely need to check / you know that $1 + \tan^2(x)$ / is going to give the square of $\sec$”)) That leads directly to $\tan^2(x) = 4$ or $\tan^2(x) = \frac{1}{4}$, giving us the same answers as previously.

I hope that helps!

- Uncle Colin