Dear Uncle Colin,

In my trigonometry homework, I’ve been asked to find the exact value of $\tan(1º)$. I have no idea where to start. Help!

-- A Nice, Gentle Little Exercise

Hello, ANGLE. That’s an evil question, on many levels. It’s doable, but it’s not at all easy. In this post, I’ll use degrees because that’s what the question asked. I don’t have to like it.

One way to approach it is:

1. Find some exact values that are 3º apart (15º and 18º are doable).

2. Work out $T = \tan(3º)$ using the angle subtraction formula

3. Expand $\tan(3x)$ in terms of $\tan(x)$ and rearrange to get a cubic in $\tan(x)$

4. Solve the cubic using Cardan’s method ((discovered by Tartaglia))

Shall we? Let’s.

$\tan(18º)$ has an exact value: $\sqrt{1 - \frac{2}{\sqrt{5}}}$, which I’ll call $T_{18}$.

$\tan(15º)$ is even less nasty: it’s $2 - \sqrt{3}$, which I’ll call $T_{15}$.

Which, when you put it into $\tan(A-B) = \frac{\tan(A)-\tan(B)}{1+ \tan(A)\tan(B)}$, gives you …

$\frac{(\sqrt{5}-1)(1+\sqrt{3})+(\sqrt{2}- \sqrt{6}) \sqrt{ 5+\sqrt{5}}}{ (\sqrt{3}-1) (\sqrt{5}-1) + (\sqrt{2}+\sqrt{6}) \sqrt{5+\sqrt{5}}}$.

Lovely. Let’s call that $T$, and say that $t = \tan(1º)$.

Now, if you expand $\tan(3x)$, you get, after a bit of algebra (or using a neat consequence of De Moivre’s theorem), $T = \frac{3t - t^3}{1-3t^2}$

That gives us the cubic $t^3 - 3T t^2 - 3t + T = 0$. What does Dr Cardan say?

Well, by Cardan, I mean Wolfram Alpha, which spits out a thing of great ugliness. However, you can tidy it up to be:

$t = \sqrt[3]{(T+\i)(T^2 + 1)} + \frac{T^2 + 1}{\sqrt[3]{(T+\i)(T^2+1)}} + T$. Those $i$s in there? Put aside mental tortures. They cancel out.

All you need to do now is replace all of the $T$s with the expression from earlier and there you go! An explicit expression for $\tan(1º)$ in terms of square and cube roots. (Personally, I’d go with $\frac{\pi}{180}$ as a pretty good approximation, but sometimes you need these things accurately.)

-- Uncle Colin.