# A lovely trigonometric identity

When I was researching the recent Mathematical Ninja piece for Relatively Prime, I stumbled on something at MathWorld ((Equation (49) and onwards)) I’d never noticed before: if $t = \tan(x)$, then:

$\tan(2x) = \frac{2t}{1 - t^2}$ $\tan(3x) = \frac{3t - t^3}{1 - 3t^2}$ $\tan(4x) = \frac{4t - 4t^3}{1 - 6t^2 + t^4}$ $\tan(5x) = \frac{5t - 10t^3 + t^5}{1 - 10t^2 + 5t^4}$

### Notice a pattern?

In case the answer was ‘no’, I’m going to pick apart the three pieces of each term: the sign, the powers of $t$ and the coefficient.

The signs in that last case are: $\frac{ + … - … +}{+ … - … +}$. That’s simple enough: in fact, they always alternate on the top and the bottom, starting from + in each case.

The powers of $t$ are fairly simple, too: $\frac{ t … t^3 … t^5 }{1 … t^2 … t^4}$. Odd powers increasing on top, even powers increasing on the bottom.

The coefficients are the neat thing, though: $\frac{ … 5 … 10 … 1}{1… 10 … 5 …}$. That’s a row from Pascal’s Triangle!

### Why should this be so?

I think it would make a good proof by induction question to show that $\tan(nx) \equiv \frac{ \nCr{n}{1} t - \nCr{n}{3} t^3 + … + (-1)^{k} \nCr{n}{2k+1} t^{2k+1} + …} {1 - \nCr{n}{2} t^2 + … + (-1)^{k} \nCr{n}{2k} t^{2k} + …} $ for all $n>2$, but I’ve not tried it and it might well be horrendous.

Instead, I’m going to appeal to complex numbers. Consider $z = 1 + t\i$. The argument of this is clearly $\arctan(t)$, which is the $x$ from before; $\tan(nx)$ is the argument of $z^n$.

Now, $(1+t \i)^n = 1 + \nCr{n}{1} t \i - \nCr{n}{2} t^2 - \nCr{n}{3} t^3 \i + …$.

The real part of that – $r \cos(nx)$ – is $1 - \nCr{n}{2} t^2 + \nCr{n}{4} t^4 - …$ : all the stuff from the bottom of the fraction.

The imaginary part, $r \sin(nx)$ is $\nCr{n}{1} t - \nCr{n}{3} t^3 + …$ : all the stuff from the top.

The tangent of the argument argument is simply the first of those divided by the second, which is what we had to begin with. How lovely is that?