Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
add comment

5 Answers

up vote 27 down vote accepted

$$\tan^{-1}(2)+\tan^{-1}(3)=\tan^{-1}{\left(\frac{2+3}{1-2\cdot 3}\right)}=\tan^{-1}(-1)=n\pi-\frac \pi 4,$$ 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 $0\neq \pi$.

share|improve this answer
    
add comment

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

share|improve this answer
2  
+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. –  Emmad Kareem Sep 16 '12 at 7:51
1  
@EmmadKareem: Sorry, it should be B, O, D are collinear, and it is pretty obvious that they all fall on the same line $y = -3x$. –  KennyTM Sep 16 '12 at 8:06
2  
Beautiful. Well done. –  bubba Sep 16 '12 at 10:05
2  
Can I ask how you made that picture? It looks really intuitive. :) –  David Sep 16 '12 at 15:38
1  
@David: GeoGebra. –  KennyTM Sep 16 '12 at 16:05
show 1 more comment

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

share|improve this answer
add comment

Proof without word

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

share|improve this answer
2  
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! –  dtldarek Jan 7 '13 at 19:30
1  
Agree with dtldarek, and trying to fix the upvote count. –  Jyrki Lahtonen Jan 7 '13 at 21:36
add comment

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.

share|improve this answer
1  
How did you manage the first equation? –  Nastassja Sep 16 '12 at 7:09
2  
I asked Maple for integer solutions of $$\frac{a+b+c-abc}{1-ab-bc-ca} = d$$ –  Robert Israel Sep 16 '12 at 7:24
    
Thanks! ${}{}{}$ –  Nastassja Sep 16 '12 at 7:25
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.