Page:Annales de mathématiques pures et appliquées, 1821-1822, Tome 12.djvu/284

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QUADRILATÈRE

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

Le numérateur de la première fraction se décompose en ces deux facteurs

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

et le numérateur de la seconde en ces deux-ci :

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

de sorte qu’on a

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

or, on a

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

donc finalement