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Why does $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$?
How to prove $$\arctan(1)+\arctan(2)+ \arctan(3)=\pi$$
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$\begingroup$ See also math.stackexchange.com/questions/197393/… $\endgroup$– Robert IsraelJan 7, 2013 at 16:13
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3 Answers
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Once I have seen a very nice proof of this: your claim is equivalent to proving that the sum of red, green and blue angles is $\pi$. Note that in the second picture, the blue-green triangle is right and isosceles (that is 45-45-90 triangle and thus similar to small red-black triangle in the first diagram).
Cheers!
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1$\begingroup$ How does one know that the red angle in the second diagram is the same as the red angle in the first? $\endgroup$ Jan 7, 2013 at 17:56
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1$\begingroup$ @StevenStadnicki "Note that on the second picture, the blue-green triangle is right and isosceles." The red-black triangle in the first diagram is also right and isosceles. Using different words, both angles are between a side and a diagonal of a square -- in the first diagram it is a black square and red diagonal, in the second it is a green square (out of four green segments only two are drawn) and blue diagonal. $\endgroup$– dtldarekJan 7, 2013 at 19:21
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$\begingroup$ Ahh, I see - I hadn't parsed that as the green-green-blue triangle being similar to the triangle the red arc describes in the top picture. Maybe a small note about how both are 45-45-90 and thus similar? That's a bit more quickly-parsed at least for me than 'right and isosceles'... $\endgroup$ Jan 7, 2013 at 19:54
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$\begingroup$ @StevenStadnicki Should be more explicit now ;-) $\endgroup$– dtldarekJan 7, 2013 at 21:53
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$\begingroup$ It is indeed, and excellent! A well-earned +1. $\endgroup$ Jan 7, 2013 at 22:22
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Consider $(1+i)$, $(1+2i)$ and $(1+3i)$ which have respective arguments of $\arctan 1$, $\arctan 2$ and $\arctan 3$. Their product equals the sum of the arguments. Thus
$$\arctan 1+\arctan 2+\arctan 3 = \operatorname {Arg} ((1+i)(1+2i)(1+3i)) = \operatorname {Arg} (-10) = \pi$$
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As $$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B},$$ $$\tan (A+B+C)=\frac{\tan(A+B)+\tan C}{1-\tan(A+B)\tan C}=\frac{\frac{\tan A+\tan B}{1-\tan A\tan B}+\tan C}{1-\tan C\left(\frac{\tan A+\tan B}{1-\tan A\tan B}\right)}=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan C\tan A}$$
So, $$A+B+C=\arctan\left(\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan C\tan A}\right)+m\pi$$ where $m$ is any integer.
Putting $\tan A=1, \tan B=2,\tan C=3$,
So, $$\arctan 1+m_1\pi+\arctan 2y+m_2\pi+\arctan 3+m_3\pi=\arctan 0+m\pi$$ where $m_i$ are integers.
The general value of $\arctan 1+\arctan 2+\arctan 3=\arctan 0+(m-m_1-m_2-m_3)\pi$ $=(n+m-m_1-m_2-m_3)\pi=r\pi$ where $n$ is any integer, hence $r=n+m-m_1-m_2-m_3$ is.
As the special value of each of the inverse trigonometric function lies in $\left(0,\frac\pi2\right)$ so their sum will lie in $(0,\frac{3\pi}2).$ Hence, the special value being $0\cdot\pi=0$
answered Jan 7, 2013 at 15:57
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$\begingroup$ I'm not convinced this will make it more clear for the OP. I suggest adding that what you did is find $\tan(\alpha+\beta+\gamma)$. $\endgroup$ Jan 7, 2013 at 16:02
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$\begingroup$ The problem is, $\tan(x) = y$ doesn't mean $\arctan(y) = x$, unless you're using a multivalued version of arctan. For the standard arctan, which has $-\pi/2 < \arctan(t) < \pi/2$ for real $t$, you have to say $\arctan(y) = x + n \pi$ for some integer $n$. You then want to use some numerical estimates to find out which $n$ it is. $\endgroup$ Jan 7, 2013 at 16:16
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$\begingroup$ +1 for the extra work. Though, didn't you notice that you've yourself answered this question before? I guess it's not hard to forget about one answer when you have 600 of them... $\endgroup$ Jan 7, 2013 at 16:16
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$\begingroup$ @KarolisJuodelė, wish we had an automatic duplicate checking :) $\endgroup$ Jan 7, 2013 at 16:48