Page:Annales de mathématiques pures et appliquées, 1821-1822, Tome 12.djvu/285

277

INSCRIT.

${\begin{aligned}&\operatorname {Sin} .^{2}{\frac {1}{2}}(a,d)=\\&{\frac {\operatorname {Cos} .{\frac {1}{4}}(a+b+c+d)\operatorname {Cos} .{\frac {1}{4}}(a+c-b-d)\operatorname {Sin} .{\frac {1}{4}}(a+b+c-d)\operatorname {Sin} .{\frac {1}{4}}(b+c+d-a)}{\left(\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d\right)\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .{\frac {1}{2}}d}},\\\\&\operatorname {Cos} .^{2}{\frac {1}{2}}(a,d)=\\&{\frac {\operatorname {Cos} .{\frac {1}{4}}(a+b-c-d)\operatorname {Cos} .{\frac {1}{4}}(b+c-a-d)\operatorname {Sin} .{\frac {1}{4}}(d+a+b-c)\operatorname {Sin} .{\frac {1}{4}}(c+d+a-b)}{\left(\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d\right)\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .{\frac {1}{2}}d}}.\\\\\end{aligned}}$

En faisant, pour abréger,

${\begin{aligned}&\operatorname {Cos} .{\frac {1}{4}}(a+b+c+d)\operatorname {Cos} .{\frac {1}{4}}(a+b-c-d)\operatorname {Cos} .{\frac {1}{4}}(a+c-b-d)\operatorname {Cos} .{\frac {1}{4}}(b+c-a-d)=M\,;\\\\&\operatorname {Sin} .{\frac {1}{4}}(b+c+d-a)\operatorname {Sin} .{\frac {1}{4}}(c+d+a-b)\operatorname {Sin} .{\frac {1}{4}}(d+a+b-c)\operatorname {Sin} .{\frac {1}{4}}(a+b+c-d)=N\,;\end{aligned}}$

auquel cas $M$ et $N$ seront des fonctions symétriques des quatre côtés $a,b,c,d,$ et en se rappelant que

$\operatorname {Sin} .(a,d)=2\operatorname {Sin} .{\frac {1}{2}}(a,d)\operatorname {Cos} .{\frac {1}{2}}(a,d),$

on aura

$\operatorname {Sin} .(a,d)={\frac {2{\sqrt {MN}}}{\left(\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d\right)\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .{\frac {1}{2}}d}}\,;\,$(III)

d’où l’on voit qu’ici les angles opposés ne sont pas supplément l’un de l’autre, comme dans le quadrilatère rectiligne inscrit.

En changeant respectivement $a,b,c,d$ en $c,d,a,b,$ il vient

${\begin{aligned}&\operatorname {Sin} .^{2}{\frac {1}{2}}(b,c)=\\&{\frac {\operatorname {Cos} .{\frac {1}{4}}(a+b-c-d)\operatorname {Cos} .{\frac {1}{4}}(a+c-b-d)\operatorname {Sin} .{\frac {1}{4}}(c+d+a-b)\operatorname {Sin} .{\frac {1}{4}}(d+a+b-c)}{\left(\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d\right)\operatorname {Cos} .{\frac {1}{2}}b\operatorname {Cos} .{\frac {1}{2}}c}},\\\\&\operatorname {Cos} .^{2}{\frac {1}{2}}(b,c)=\\&{\frac {\operatorname {Cos} .{\frac {1}{4}}(a+b+c+d)\operatorname {Cos} .{\frac {1}{4}}(b+c-a-d)\operatorname {Sin} .{\frac {1}{4}}(b+c+d-a)\operatorname {Sin} .{\frac {1}{4}}(a+b+c-d)}{\left(\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d\right)\operatorname {Cos} .{\frac {1}{2}}b\operatorname {Cos} .{\frac {1}{2}}c}}\,;\\\\\end{aligned}}$

mais on a