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?  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.\$\$ As you discussed "geometric proof", let me try to carry out one: ;) Hope it helps though...

You are watching: Arctan(-1/2) You can likewise directly use the following formula:

\$\$ ag1 an(alpha+eta+gamma) = frac analpha+ aneta+ angamma- analpha aneta angamma1- analpha aneta- analpha angamma- aneta 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)\$.