How to add four arctangents efficiently

[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

 

[janhaa]

maybe this can help you

http://www.enotes.com/homework-help/evaluate-value-this-expression-arctan-1-3-arctan-1-251223

 

[Ray Vickson]

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

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.