Page:Annales de mathématiques pures et appliquées, 1824-1825, Tome 15.djvu/303

$\operatorname {Tang} .{\frac {1}{2}}s=\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad$(xxxvi)

${\sqrt {-\operatorname {Tang} .{\frac {1}{2}}\left({\frac {1}{2}}\varpi -S\right)\operatorname {Cot} .{\frac {1}{2}}\left\{{\frac {1}{2}}\varpi -(S-A)\right\}\operatorname {Cot} .{\frac {1}{2}}\left\{{\frac {1}{2}}\varpi -(S-B)\right\}\operatorname {Cot} .{\frac {1}{2}}\left\{{\frac {1}{2}}\varpi -(S-C)\right\}}}.$

Par les mêmes formules (5), et par la formule (XXIX), on a

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

comparant ces formules aux formules (29) et (30), on pourra leur donner cette forme

$-\operatorname {Sin} .{\frac {1}{2}}\left({\frac {1}{2}}\varpi -S\right)={\sqrt {\frac {\operatorname {Sin} .{\frac {1}{2}}s\operatorname {Sin} .{\frac {1}{2}}(s-a)\operatorname {Sin} .{\frac {1}{2}}(s-b)\operatorname {Sin} .{\frac {1}{2}}(s-c)}{\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .{\frac {1}{2}}b\operatorname {Cos} .{\frac {1}{2}}c}}},$(XXXV)

$+\operatorname {Cos} .{\frac {1}{2}}\left({\frac {1}{2}}\varpi -S\right)={\sqrt {\frac {\operatorname {Cos} .{\frac {1}{2}}s\operatorname {Cos} .{\frac {1}{2}}(s-a)\operatorname {Cos} .{\frac {1}{2}}(s-b)\operatorname {Cos} .{\frac {1}{2}}(s-c)}{\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .{\frac {1}{2}}b\operatorname {Cos} .{\frac {1}{2}}c}}},$(XXXIV)

d’où encore

$-\operatorname {Tang} .{\frac {1}{2}}\left({\frac {1}{2}}\varpi -S\right)=$

${\sqrt {\operatorname {Tang} .{\frac {1}{2}}s\operatorname {Tang} .{\frac {1}{2}}(s-a)\operatorname {Tang} .{\frac {1}{2}}(s-b)\operatorname {Tang} .{\frac {1}{2}}(s-c)}}.$^[1](XXXVI)

Par les formules (5) et par la formule (xxx), on trouve

1. Si l’on désigne par $T$ l’aire du triangle, on aura, comme l’on sait,
   $T=A+B+C-\varpi =2S-\varpi ,\quad$d’où$\quad {\frac {1}{2}}T=-\left({\frac {1}{2}}\varpi -S\right),$

   et par conséquent

   $\operatorname {Sin} .{\frac {1}{2}}T=-\operatorname {Cos} .S,\quad \operatorname {Cos} .{\frac {1}{2}}T=\operatorname {Sin} .S,\quad \operatorname {Tang} .{\frac {1}{2}}T=-\operatorname {Cot} .S,$

   on aura en outre ${\frac {1}{4}}T=-{\frac {1}{2}}\left({\frac {1}{2}}\varpi -S\right),$ d’où

   $\operatorname {Sin} .{\frac {1}{4}}T=-\operatorname {Sin} .{\frac {1}{2}}\left({\frac {1}{2}}\varpi -S\right),\ \operatorname {Cos} .{\frac {1}{4}}T$
   $=\operatorname {Cos} .{\frac {1}{2}}\left({\frac {1}{2}}\varpi -S\right),\ \operatorname {Tang} .{\frac {1}{4}}T=-\operatorname {Tang} .{\frac {1}{2}}\left({\frac {1}{2}}\varpi -S\right),$