# How to add four arctangents efficiently

1. May 3, 2013

### mgamito

I need to compute the addition of four arctangents:

$\alpha = \arctan(x_1) + \arctan(x_2) + \arctan(x_3) + \arctan(x_4)$

Rather than call four arctangent functions, I was thinking of using the arctangent addition formula:

$\arctan(u) + \arctan(v) = \arctan(\frac{u + v}{1 - uv}) + \pi n$, for some $n \in N$

I could invoke the above property three times and I would only have to call the arctan function once. My problem is how to keep track of the integer $n$ throughout so that the result is correct.

My initial variables $x_1$ to $x_4$ are all positive so the result should be an angle $0 \leq \alpha < 2\pi$. I suspect the answer lies in keeping track of the sign of the $1 - uv$ denominators and introducing some factor of $\pi$ correction if they go negative.

I'll work this through the weekend but I thought I would post this here in case someone figured this out already.

Thank you,
manuel

2. May 3, 2013

3. May 3, 2013

### Ray Vickson

If $\arctan(x_1) + \arctan(x_2) = \arctan(y_1) + \pi n_1$ and $\arctan(x_3) + \arctan(x_4) = \arctan(y_2) + \pi n_2,$ we then have $\sum_{i=1}^4 \arctan(x_i) = \arctan(y_1) + \arctan(y_2) + \pi (n_1 + n_2),$ which has the form $\arctan(z) + \pi (n_1 + n_2 + n_3).$ Here,
$$z = \frac{y_1 + y_2}{1-y_1 y_2}.$$
We might as well just say that $n_1 + n_2 + n_3 = n,$ an integer.

Last edited: May 3, 2013