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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.)

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This seems scattered. What is $y$ here? – Cameron Williams Aug 12 '13 at 20:48
I'm not sure it helps, but you can "split" the quotient inside the arctan like this: $\arctan({\frac{x}{c}}) = \arctan{(\frac{x + \sqrt{x^2 + 4(c-1)}}{2})} + \arctan{(\frac{x - \sqrt{x^2 + 4(c-1)}}{2})}$ – Daniel R Aug 12 '13 at 20:51
@CameronWilliams: Sorry. I rewrote the question and used c instead of the original y to show that it is a constant. – yzt Aug 12 '13 at 21:03
@DanielR: Ideally, I want to get rid of the $\arctan$ altogether, if at all possible. – yzt Aug 12 '13 at 21:04

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:

$$\arctan \frac{x}{y} = \sum_{i=0}^\infty (-1)^i\frac{1}{2i+1}\left(\frac{x}{y}\right)^{2i+1}$$

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There is no way to remove the $arctan$ altogether, is there? – yzt Aug 12 '13 at 21:06
You can check the edit i've made – Stefan4024 Aug 12 '13 at 21:19
This is interesting and probably the way I'd have to go eventually. Thank you for your answer! Now, it seems to me that for very small $x$s relative to $y$, $\arctan \frac xy$ is pretty close to $\frac xy$ (because of the increasing powers on the subsequent $\frac xy$s.) Of course, I'd have to investigate the error bounds to see whether it is acceptable. But in you opinion, wouldn't it be easier to simply drop the $\arctan$ altogether and be rid of it? – yzt Aug 12 '13 at 21:31
If you want just an approximate answer that would do it, note that if x is 10 000 times smaller than y the value will be around 500 times greater than the value of the angle, but the difference will be $\approx$ 0.0049. You can overlook that difference. If you need strictly correct answer i wouldn't recommend this. – Stefan4024 Aug 12 '13 at 21:41

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