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I've looked at Wikipedia and don't really understand what I'm reading or how to implement atan2(). I was hoping someone could point me to a better resource or give me an example of how to use atan2().

For some context, I'm asking about atan2() because I have a a homework asking:

Write the solution for each of the following in terms of atan2() before giving your final answer (e.g., = atan2(a,b) = ...

An example problem: *If

sin $\theta$ = -1/2

and

cos $\theta$ = $\sqrt{3}/2$

what is $\theta$ ?

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    $\begingroup$ $\cos\,\theta$ is always between $-1$ and $1$ for real angles, so I can't see how it'd be equal to $3/2$. Anyway, two-argument arctangent is essentially the same as the arctangent of a certain ratio. Have you already taken up polar coordinates? $\endgroup$ – J. M. is a poor mathematician Sep 23 '11 at 18:27
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    $\begingroup$ You might want to think a bit about your example problem. The value of $\cos(\theta)$ is always between $-1$ and $+1$, and so $\cos(\theta) = 3/2$ is not right. $\endgroup$ – Dilip Sarwate Sep 23 '11 at 18:27
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    $\begingroup$ But now you have too much of a square root. It should be $\sqrt{3}/2$ -- otherwise $\sin^2+\cos^2$ doesn't hold. $\endgroup$ – Henning Makholm Sep 23 '11 at 18:36
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    $\begingroup$ @Christian, atan2 is nothing more or less than how most programming language spell the name of the principal-argument function. What's sick about that? $\endgroup$ – Henning Makholm Sep 23 '11 at 20:11
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    $\begingroup$ You will need to check the order of the arguments for atan2 in your particular system. If atan2(1,0)=0 then you want $\text{atan2}(r\sin(\theta),r\cos(\theta))$; otherwise atan2(0,1)=0 and you want $\text{atan2}(r\cos(\theta),r\sin(\theta))$. $\endgroup$ – Henry Sep 23 '11 at 21:28
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Are you asking how to use atan2, or how to implement it?

$\mathrm{atan2}(y,x)$ gives you the angle between the x-axis and the vector $(x,y)$, usually in radians and signed such that for positive $y$ you get an angle between $0$ and $\pi$, and for negative $y$ the result is between $-\pi$ and $0$.

It's named that way because when $x$ is positive, $\mathrm{atan2}(y,x)=\tan^{-1}(\frac{y}{x})$, which also explains (more or less) why the arguments are "backwards".

So if you have the vector $(\sqrt 3/2, -1/2)$ you can get its angle by $\mathrm{atan2}(-1/2, \sqrt 3/2)$ which is $-\pi/6$. You then know that for some positive $r$ it holds that $$(\sqrt 3/2, -1/2) = r(\cos(-\pi/6), \sin(-\pi/6))$$ (in fact it turns out that $r=1$ in this case, but atan2 does not tell you that).

As for implementing atan2 in the (these days somewhat unusual) case that you don't have a hardware implementation available, I think CORDIC should give excellent results with little effort. Alternatively, you can sacrifice precision for table space by splitting into octants, dividing the numerically smaller argument by the larger, and using a polynomial approximation of the arctangent between 0 and 1.

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  • $\begingroup$ Heh, I picked up something useful from the ol' wiki. You can "emulate" atan2() by picking the appropriate root $t^\ast$ of the quadratic $yt^2+2xt-y=0$ and then $\arctan(x,y)=2\arctan\,t^\ast$. One might be able to squeeze a bit more accuracy if one uses a rational approximation instead of a polynomial one. $\endgroup$ – J. M. is a poor mathematician Sep 24 '11 at 2:43
  • $\begingroup$ Sorry to come back, but does r = the angle? $\endgroup$ – Nick Sep 24 '11 at 15:07
  • $\begingroup$ No, $r$ is the absolute length of the vector. $\endgroup$ – Henning Makholm Sep 24 '11 at 15:08
  • $\begingroup$ Ah, I got -30 for the angle, does that sound right? $\endgroup$ – Nick Sep 24 '11 at 15:43
  • $\begingroup$ Yes, $-30^\circ = -\pi/6$. $\endgroup$ – Henning Makholm Sep 24 '11 at 15:46
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The slope of the straight line from the origin $(0,0)$ to the point $(a,b)$ is $b/a$ which takes on values from $-\infty$ to $+\infty$. But if you look at the angle between the line from $(0,0)$ to $(a,b)$, this angle can be anything from $0$ to $2\pi$. In particular, the line from $(0,0)$ to $(a,b)$ has the same slope as the line from $(0,0)$ to $(-a, -b)$ but the angle differs by $\pi$. The atan2 function was developed to take into account this difference. If you specify the angle as $\arctan(b/a)$ or $\text{atan}(b/a)$, you will get an answer between $-\pi/2$ and $+\pi/2$, that is, atan() cannot distinguish between the points $(a,b)$ and $(-a,-b)$ in terms of the angle. But if you specify the angle as: $\text{atan2}(a,b)$, you will get an answer between $0$ and $2\pi$ because the information about the signs of $a$ and $b$ (which tells you whether the point is in the first, second, third, or fourth quadrants) is not lost.

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