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Playing around on wolframalpha shows $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$. I know $\tan^{-1}(1)=\pi/4$, but how could you compute that $\tan^{-1}(2)+\tan^{-1}(3)=\frac{3}{4}\pi$ to get this result?

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    $\begingroup$ It is true in general that $\tan^{-1}(a) + \tan^{-1}(b) + \tan^{-1}(c) = \pi$ when $a+b+c=abc$ (and $a,b,c$ positive). This is the converse of what is proved here. $\endgroup$ Oct 19, 2015 at 11:55

6 Answers 6

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enter image description here

Consider $O=(0,0)$, $A=(1,1)$, $B=(-1,3)$, $D=(1,-3)$, $E=(1,0)$.

\begin{align} 2 &= \frac{AB}{AO} = \tan \angle AOB \\ 1 &= \frac{AE}{EO} = \tan \angle AOE \\ 3 &= \frac{DE}{DO} = \tan \angle DOE \end{align}

The points B, O and D are collinear, i.e. $\angle BOD = \tan^{-1}2+\tan^{-1}1+\tan^{-1}3 = \pi$.

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    $\begingroup$ +1, I like showing facts in drawings - Somehow it makes facts easier to grasp. It may be nicer to explain the part: "The points B, O and E are collinear,..." a bit. Thanks. $\endgroup$
    – NoChance
    Sep 16, 2012 at 7:51
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    $\begingroup$ @EmmadKareem: Sorry, it should be B, O, D are collinear, and it is pretty obvious that they all fall on the same line $y = -3x$. $\endgroup$
    – kennytm
    Sep 16, 2012 at 8:06
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    $\begingroup$ Beautiful. Well done. $\endgroup$
    – bubba
    Sep 16, 2012 at 10:05
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    $\begingroup$ Can I ask how you made that picture? It looks really intuitive. :) $\endgroup$
    – David G
    Sep 16, 2012 at 15:38
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    $\begingroup$ @David: GeoGebra. $\endgroup$
    – kennytm
    Sep 16, 2012 at 16:05
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Proof without word

$\tan^{-1} 1+\tan^{-1} 2+\tan^{-1} 3 =\pi$.

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    $\begingroup$ Oh! I answered a duplicated question and here is the same nice solution (compare this). I don't know why this has only 4 upvotes :-| +1! $\endgroup$
    – dtldarek
    Jan 7, 2013 at 19:30
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    $\begingroup$ Agree with dtldarek, and trying to fix the upvote count. $\endgroup$ Jan 7, 2013 at 21:36
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    $\begingroup$ How do you do geometric Proofs? $\endgroup$
    – user427802
    Jul 27, 2018 at 16:06
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The simplest way is by using complex numbers. It is a trivial computation to show that $$(1+i)(1+2i)(1+3i)=-10$$ Now recall the geometric description of complex multiplication (multiply the lengths and add the angles), and take the argument on both sides of this equation. This gives $$\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$$

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$$\tan^{-1}(2)+\tan^{-1}(3)=\tan^{-1}{\left(\frac{2+3}{1-2\cdot 3}\right)}=\tan^{-1}(-1)=n\pi-\frac \pi 4,$$ where $n$ is any integer.

Now the principal value of $\tan^{-1}(x)$ lies in $[-\frac \pi 2, \frac \pi 2]$ precisely in $(0, \frac \pi 2)$ if finite $x>0$. So, the principal value of $\tan^{-1}(2)+\tan^{-1}(3)$ will lie in $(0, \pi) $.

So, the principal value of $\tan^{-1}(2)+\tan^{-1}(3)$ will be $\frac {3\pi} 4$.

Interestingly, the principal value of $\tan^{-1}(-1)$ is $-\frac {\pi} 4$.

But the general values of $\tan^{-1}(2)+\tan^{-1}(3)$ and $\tan^{-1}(-1)$ are same.

Alternatively, $$\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\tan^{-1}{\left(\frac{1+2+3-1\cdot 2\cdot 3}{1-1\cdot 2- 2\cdot 3 -3\cdot 1}\right)}=\tan^{-1}(0)=m\pi$$, where $m$ is any integer.

Now the principal value of $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)$ will lie in $(0 ,\frac {3\pi} 2)$ which is $\pi$.

The principal value of $\tan^{-1}(0)$ is $0\neq \pi$.

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  • $\begingroup$ But typing the value in calculator gives π and not 0. $\endgroup$
    – suiz
    Apr 15, 2018 at 11:10
  • $\begingroup$ @suiz, $$\tan^{-1}(2)+\tan^{-1}(3)$$ will be $$\frac {3\pi} 4$$ right? $\endgroup$ Apr 15, 2018 at 13:00
  • $\begingroup$ Yes. And arctan(1) gives π/4. If we add it with 3π/4, we get π. Am I wrong with this? $\endgroup$
    – suiz
    Apr 15, 2018 at 17:24
  • $\begingroup$ @suiz, You are correct. See also: math.stackexchange.com/questions/523625/… $\endgroup$ Apr 15, 2018 at 17:28
  • $\begingroup$ Also, xy >0 in this case. Then how could your writing the value 0 be right? Am I missing something out? $\endgroup$
    – suiz
    Apr 15, 2018 at 17:30
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Note that $$ \tan \left(\arctan(1+z) + \arctan\left(2 + z + z^2 \right) + \arctan \left( 3+3\,z+4\,{z}^{2}+2\,{z}^{3}+{z}^{4} \right) \right)=z $$ so that $$\arctan(1+z) + \arctan\left(2 + z + z^2 \right) + \arctan \left( 3+3\,z+4\,{z}^{2}+2\,{z}^{3}+{z}^{4} \right) = \arctan(z) + n \pi $$ for the appropriate integer $n$. For integers $z$ we get interesting arctan identities from this.

$$\eqalign{ \arctan(1) + \arctan\left(2\right)+ \arctan\left(3\right) &= \pi \cr \arctan(2) + \arctan(4) + \arctan(13) &= \arctan(1) + \pi \cr \arctan(3) + \arctan(8) + \arctan(57) &= \arctan(2) + \pi \cr \arctan(4) + \arctan(14) + \arctan(183) &= \arctan(3) + \pi \cr}$$ etc.

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    $\begingroup$ How did you manage the first equation? $\endgroup$
    – Nastassja
    Sep 16, 2012 at 7:09
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    $\begingroup$ I asked Maple for integer solutions of $$\frac{a+b+c-abc}{1-ab-bc-ca} = d$$ $\endgroup$ Sep 16, 2012 at 7:24
  • $\begingroup$ I know that this an old answer, but would you please explain how the integer solutions of $$\frac{a+b+c-abc}{1-ab-bc-ca} = d$$ relate to your first identity? I can't see it on my own :( $\endgroup$
    – math.n00b
    Jul 23, 2014 at 9:58
  • $\begingroup$ Maple gave me a parametric solution which simplifies to $$\left\{ a={{\it \_Z1}}^{4}+2\,{{\it \_Z1}}^{3}+4\,{{\it \_Z1}}^{2}+3 \,{\it \_Z1}+3,b={{\it \_Z1}}^{2}+{\it \_Z1}+2,c=1+{\it \_Z1},d={\it \_Z1} \right\} $$ $\endgroup$ Jul 23, 2014 at 15:12
  • $\begingroup$ @math.n00b $\tan(\arctan(a)+\arctan(b)+\arctan(c)) = ?$ $\endgroup$
    – Max Payne
    Nov 5, 2016 at 7:18
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Consider, $z_1= \frac{1+2i}{\sqrt{5}}$, $z_2= \frac{1+3i}{\sqrt{10} }$, and $z_3= \frac{1+i}{\sqrt{2} }$, then:

$$ z_1 z_2 z_3 = \frac{1}{10} (1+2i)(1+3i)(1+i)=-1 $$

Take arg of both sides and use property that $\arg(z_1 z_2 z_3) = \arg(z_1) + \arg(z_2) + \arg(z_3)$:

$$ \arg(z_1) + \arg(z_2) + \arg(z_3) = -1$$

The LHS we can write as:

$$ \tan^{-1} ( \frac{2}{1}) +\tan^{-1} ( \frac{3}{1} ) + \tan^{-1} (1) = \pi$$

Tl;dr: Complex number multiplication corresponds to tangent angle addition

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