Page:Annales de mathématiques pures et appliquées, 1821-1822, Tome 12.djvu/286

278

QUADRILATÈRE

 Les corrections sont expliquées en page de discussion

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

en substituant donc, et posant, pour abréger,

$\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .{\frac {1}{2}}b\operatorname {Cos} .{\frac {1}{2}}c\operatorname {Cos} .{\frac {1}{2}}d=P,$

il viendra

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

cela revient à

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

d’où, par un développement ultérieur,

$\operatorname {Sin} .{\frac {1}{2}}\left\{(a,d)+(b,c)\right\}=+{\sqrt {\frac {M}{P}}},\quad \operatorname {Cos} .{\frac {1}{2}}\left\{(a,d)+(b,c)\right\}=-{\sqrt {\frac {N}{P}}}.$

Ces deux fonctions étant symétriques, il en résulte que

$\operatorname {Sin} .{\frac {1}{2}}\left\{(a,d)+(b,c)\right\}=\operatorname {Sin} .{\frac {1}{2}}\left\{(a,b)+(c,d)\right\}=+{\sqrt {\frac {M}{P}}},$