Where in the angles we are searching for are displayed, however I can"t fix it. Can we use it with this type of approach? Can someone additionally article a solution making use of 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$$ arctanfrac12+arctanfrac15+arctanfrac18=arctan 1=fracpi4.$$


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As you discussed "geometric proof", let me try to carry out one: ;)

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Hope it helps though...

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You can likewise directly use the following formula:

$$ ag1 an(alpha+eta+gamma) = frac analpha+ aneta+ angamma- analpha aneta angamma1- analpha aneta- analpha angamma- aneta angamma$$

Taking $arctan$ of both sides, and also setting

$$a:= an alpha, b:= an alpha, c := an gamma,$$

we obtain:

$$ ag2arctan(a)+arctan(b)+arctan(c)=arctan left( fraca+b+c-abc1-ab-ac-bc ight)$$

It continues to be to rearea $a,b,c$ by their values to obtain

$$arctan 1=dfracpi4$$

Renote 1 : A doprimary of validity of formula (1) is for angles $alpha, eta, gamma in (0, pi/2)$ such that $alpha+eta+gamma in (0, pi/2)$ too. Here, these conditions are fulfilled.

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

It is an prompt consequence of the adhering to identification in $ugandan-news.combbC$:

$$ ag3(1+ia)(1+ib)(1+ic)=1+i(a+b+c)+i^2(ab+ac+bc)+i^3 abc$$

Due to the fact that, taking arguments on both sides of (3), under the condition given in Renote 1 (that avoid including $+k2pi$ or $+kpi$):

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

which is nopoint else than (2).

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Remark 2: on the model of (2), one can expush a sum of $arctan$ of any dimension under a closed develop $arctan(cdots)$.