I don't know enough trigonometry and math to do what I'm trying to do. Any help is appreciated.
I have a formula like this:
$$a = \arctan \left( \frac xc \right)$$
and what I'm trying to do is convert the above equation into something like this:
$$a = \frac pq,$$
$$ \begin{bmatrix}p \\ q \end{bmatrix} = \begin{bmatrix}m_{11} & m_{12} \\ m_{21} & m_{22} \end{bmatrix} \begin{bmatrix}x \\ c \end{bmatrix}$$
In other words, I want to find the four matrix elements such that
$$ \arctan \left( \frac xc \right) = \frac {m_{11} \times x + m_{12} \times c}{m_{21} \times x + m_{22} \times c} $$
If it helps, I know that $0 < x < c$. If fact, $x$ is much smaller than $c$, but not so small that $\frac xc$ can be considered $0$.
Is this possible? Even an approximation (within certain bounds) would be fine.
(Note: I edited the question heavily in order to better describe what I'm trying to achieve. This will make some comments and even answers seem weird and irrelevant, which they weren't before I did the edit. For this, I apologize.)
Best Answer
You can do this. Replace $x = a + b$ and $y = 1 - ab$. Now we can get:
We can modify this to: $-x = -(a+b)$ and $-(y-1) = ab$.
Using Vieta's formula we can get solution for $a$ and $b$ solving this quadratic eqation:
$$t^2 - xt - y + 1 = 0$$
Now we can use this trigonometric identity:
$$\arctan (a) + \arctan (b) = \arctan \left( \frac{a+b}{1-ab} \right) = \arctan \left(\frac{x}{y}\right)$$
In my opinion this is as much as you can go.
You can't just remove the $arctan$, but you can get it value for $\arctan \frac{x}{y}$ using this formula:
$$arctan \frac{x}{y} \approx \frac{x}{y} - \frac{1}{3}\left(\frac{x}{y}\right)^3 + \frac{1}{5}\left(\frac{x}{y}\right)^5...$$
This is Taylor-Maclaurin series and you'll get closer and closer to the true value with each step, although you will not get exactly the true falue, but you'll be close enough.
The summation formula for the series is:
$$\arctan \frac{x}{y} = \sum_{i=0}^\infty (-1)^i\frac{1}{2i+1}\left(\frac{x}{y}\right)^{2i+1}$$