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How do you solve $\arctan(n)$ to radians by hand? I. e. $\arctan(1)$ >> process >> $\pi/4$

I have this Taylor expansion that allows me to calculate an approximate value for arctan, but am wondering if there's a closed-form solution (Or a more general formula than below):

$$\begin{align}\arctan z&=z-\frac{z^3}3+\frac{z^5}5-\frac{z^7}7+\cdots\\&=\sum_{n=0}^\infty\frac{(-1)^nz^{2n+1}}{2n+1};\quad\|z\|\leq1\quad z\ne i,-i\end{align}$$

ACB
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Mr. Demetrius Michael
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    There are infinite series for arctan, however, this is not very convergent around 1.$$\arctan x = \sum^{\infty}_{n=0} \frac{(-1)^n}{2n+1} x^{2n+1}\quad\text{ for }|x| \le 1!$$ If you are using this method to try and find $\pi$, there are many more efficient ways. – Argon Apr 05 '12 at 21:33
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    If you really are the sort of person that computes by hand, $\arctan(1) = \pi/4$ should be something you have memorized. (or something equivalent to it, such as the isosceles right triangle) –  Apr 05 '12 at 21:35
  • Are you asking for a "closed form"? Or an approximation? – Aryabhata Apr 05 '12 at 21:40
  • Closed form, general solution. Will be programming this, as you can see, I've already memorized what ArcTan(1) is :P. I'm asking because I use mathematica, and it gives me ArcTan(2) as the answer to ArcTan(2), and irrational numbers regardless of the precision, won't do me any good. – Mr. Demetrius Michael Apr 05 '12 at 21:48
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    For small angles, $\arctan x \approx x$. Most arctangents that you take will result in irrational numbers. In a programming environment, normally a floating point answer that is close enough to the truth is the best you will do. But in most of those environments, there is already a function available. So what are you trying to do? – Ross Millikan Apr 05 '12 at 22:11
  • I one-upped your comment Argon. and edited my Original Q. – Mr. Demetrius Michael Apr 05 '12 at 22:14
  • @RossMillikan Just looking for a closed form solution, if it currently exists. – Mr. Demetrius Michael Apr 05 '12 at 22:17
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    The closed form solution is $\arctan(x)$. There is no pleasant closed form using real functions, unless you allow $\arcsin$ or $\arccos$ as a component. One can get something involving complex logarithms, but I do not think that you would consider the result satisfactory. – André Nicolas Apr 05 '12 at 22:18
  • @AndréNicolas referring to: http://en.wikipedia.org/wiki/Arctan#Indefinite_integrals_of_inverse_trigonometric_functions ? – Mr. Demetrius Michael Apr 05 '12 at 22:21
  • Didn't see it at the place you linked to. But recall that we get $\arctan x$ by integrating $\frac{1}{1+x^2}$. If we use complex numbers, we can express $\frac{1}{1+x^2}$ as $(1/2)(\frac{1}{1+ix}+\frac{1}{1-ix})$, and when we integrate we get complex logarithms. – André Nicolas Apr 05 '12 at 22:31
  • The only rational numbers $r$ for which $\arctan(r)$ is a rational multiple of $\pi$ are $\pm 1$ and $0$. See e.g. http://math.stackexchange.com/questions/94478/sin-1-circ-is-irrational-but-how-do-i-prove-it-in-a-slick-way-and-tan1/94488#94488 – Robert Israel Apr 05 '12 at 22:36
  • @Demetrius I don't know if this helps, but you can get the decimal value for an exact expression in Mathematica using the N function. Thus N[ArcTan[2]] will give the decimal value for the arctangent of 2. – Jim Belk Apr 06 '12 at 01:26
  • Thanks jim. Already know that with mathematica. I guess the real answer is that the closed form solution for ArcTan hasn't been discovered yet, but there are ways to approx it. – Mr. Demetrius Michael Apr 09 '12 at 01:08

2 Answers2

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With a good understanding, you ought to be able to figure out $\mathrm{arctan}(1)$ in your head. Here's what you need to understand:

  1. The tangent function converts angles to slopes. For example, $\mathrm{tan}(37^\circ)$ gives the slope of a line that makes an angle of $37^\circ$ with the $x$-axis.

  2. Therefore, the inverse function (arctangent) converts slopes to angles. For example, $\mathrm{arctan}(2)$ would give the angle between the line $y=2x$ and the $x$-axis.

  3. A line with slope $1$ is inclined at a $45^\circ$ angle. Therefore, $\mathrm{arctan}(1) = 45^\circ$. Converting to radians gives $\mathrm{arctan}(1)=\pi/4$.

Finding the exact arctangent of other values would be much more complicated, though you ought to be able to estimate the arctangent by picturing it. For example, it's easy to estimate that $\mathrm{arctan}(1/3)$ should be about $15$ or $20$ degrees, just by picturing a line with slope $1/3$.

Edit: By the way, if you really want to compute arctangents by hand, one possible method is to use the identity $$ \arctan(x) \;=\; 2\arctan\biggl(\frac{x}{1+\sqrt{1+x^2}}\biggr), $$ which follows from the double-angle formula for tangent. The quantity in parentheses on the right is less than $x/2$, so you can iterate this identity to find a sequence of smaller and smaller angles whose arctangents you want to figure out. (Note that you need to be able to compute square roots by hand.) Once your angle gets small enough, the approximation $$ \arctan(x) \;\approx\; x $$ becomes very accurate.

Jim Belk
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$\arccos\left(\frac1{\sqrt{1+x^2}}\right)$ is an alternate and an easier function to deal with when using $\arctan(x)$, since you can remember the Taylor series for cosine quite easily.

Take $1-\frac{x^2}2 + \frac{x^4}{24} -\ldots$ and plug in $\frac1{\sqrt{1+x^2}}$ for $x$, then set this equal to whatever you want to take the $\arctan$ of, and solve for $x$.

P.S.: this is very labor intensive

Edit: changed wording

PM 2Ring
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Luke CHILTON
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