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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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    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](http://math.stackexchange.com/questions/477364/prove-that-tan-a-tan-b-tan-c-tan-a-tan-b-tan-c-abc-180-circ).2015-10-19

5 Answers 5

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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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    @math.n00b $\tan(\arctan(a)+\arctan(b)+\arctan(c)) = ?$2016-11-05
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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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    @kennytm Only if you switch the $\tan^{-1}1$ and $\tan^{-1}2$ around.2015-04-13
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Proof without word

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

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    How do you do geometric Proofs?2018-07-27
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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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    @suiz, I think you have meant $(0,\pi)$ right? Then $($ mean open interval2018-04-15