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What is the value of : $$\arctan(1/2)+\arctan(1/5)+\arctan(1/8)?$$

I tried to do geometric solution:: enter image description here

Where in the angles we are looking for are shown, but I can't solve it. Can we use it with this kind of approach? Can someone also post a solution using trigonometric identities?

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By considering that $$(2+i)(5+i)(8+i) = 65(1+i)$$ and by taking the argument of both sides we immediately have $$ \arctan\frac{1}{2}+\arctan\frac{1}{5}+\arctan\frac{1}{8}=\arctan 1=\frac{\pi}{4}.$$

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  • $\begingroup$ [+1] Factorisation of Gaussian integers... $\endgroup$ – Jean Marie Dec 5 '17 at 9:17
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As you mentioned "geometric proof", let me try to provide one: ;)

why is this 45 degree

Hope it helps though...

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  • $\begingroup$ [+1] I tried... but I did not succed... Besides have you seen the solution by Jack d'Aurizio ? $\endgroup$ – Jean Marie Dec 5 '17 at 9:20
  • $\begingroup$ Well yes. Of course to prove this the standard method is those by Jack d'Aurizio and Robert Z, and this "geometric proof" is hard to extend, especially if you do not have standard arguments in mind. Essentially Jack's proof provides a geometric proof as well, with points on the coordinates (0,0), (80, 10), (78, 26) and (65, 65), and what I did is simply an attempt to squeeze everything in smaller coordinates. $\endgroup$ – Hw Chu Dec 7 '17 at 3:37
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Hint. Note that if $|\arctan x+\arctan y|<\pi/2$, then $$\arctan{x}+\arctan{y}=\arctan\left(\frac{x+y}{1-xy}\right).$$ (see for example HERE). Hence $$\arctan(1/5)+\arctan(1/8)=\arctan\left(\frac{1/5+1/8}{1-1/40}\right)=\arctan\left(1/3\right).$$ Can you take it from here?

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  • $\begingroup$ yes i got 45 already but where can u find the other inverse formulas? $\endgroup$ – SuperMage1 Dec 4 '17 at 10:16
  • $\begingroup$ oh wait u already linked it, thanks $\endgroup$ – SuperMage1 Dec 4 '17 at 10:17
  • $\begingroup$ Yes, see the link. For the $\arctan$ see also the addition formula for $\tan$. $\endgroup$ – Robert Z Dec 4 '17 at 10:19
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You can also directly use the following formula:

$$\tag{1}\tan(\alpha+\beta+\gamma) = \frac{\tan\alpha+\tan\beta+\tan\gamma-\tan\alpha\tan\beta\tan\gamma}{1-\tan\alpha\tan\beta-\tan\alpha\tan\gamma-\tan\beta\tan\gamma}$$

Taking $\arctan$ of both sides, and setting

$$a:=\tan \alpha, b:=\tan \alpha, c :=\tan \gamma,$$

we obtain:

$$\tag{2}\arctan(a)+\arctan(b)+\arctan(c)=\arctan \left( \frac{a+b+c-abc}{1-ab-ac-bc}\right)$$

It remains to replace $a,b,c$ by their values to obtain

$$\arctan 1=\dfrac{\pi}{4}$$

Remark 1 : A domain of validity of formula (1) is for angles $\alpha, \beta, \gamma \in (0, \pi/2)$ such that $\alpha+\beta+\gamma \in (0, \pi/2)$ as well. Here, these conditions are fulfilled.

Proof of formula (2): (that will explain the presence in (2) of symmetric polynomials $1, \ a+b+c, \ ab+ac+bc,\ abc$).

It is an immediate consequence of the following identity in $\mathbb{C}$:

$$\tag{3}(1+ia)(1+ib)(1+ic)=1+i(a+b+c)+i^2(ab+ac+bc)+i^3 abc$$

Because, taking arguments on both sides of (3), under the condition given in Remark 1 (that avoid adding $+k2\pi$ or $+k\pi$):

$$\arg(1+ia)+\arg(1+ib)+\arg(1+ic)=\arg(1-(ab+ac+bc))+i(a+b+c-abc)$$

which is nothing else than (2).

Remark 2: on the model of (2), one can express a sum of $\arctan$ of any size under a closed form $\arctan(\cdots)$.

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