Page:Annales de mathématiques pures et appliquées, 1821-1822, Tome 12.djvu/282

274

QUADRILATÈRE

 Les corrections sont expliquées en page de discussion

$\left.{\begin{aligned}\operatorname {Sin} .{\frac {1}{2}}x\operatorname {Sin} .{\frac {1}{2}}y&=\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}d,\\\\{\frac {\operatorname {Sin} .{\frac {1}{2}}x}{\operatorname {Sin} .{\frac {1}{2}}y}}&={\frac {\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c\operatorname {Sin} .{\frac {1}{2}}d}{\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d}}\end{aligned}}\right\}$(II)

Dans le triangle sphérique dont les trois côtés sont $a,d,x,$ on a

$\operatorname {Cos} .(a,d)={\frac {\operatorname {Cos} .x-\operatorname {Cos} .a\operatorname {Cos} .d}{\operatorname {Sin} .a\operatorname {Sin} .d}},$

d’où

${\begin{array}{rll}2\operatorname {Sin} .^{2}{\frac {1}{2}}(a,d)&=1-\operatorname {Cos} .(a,d)&={\frac {\operatorname {Cos} .(a-d)-\operatorname {Cos} .x}{\operatorname {Sin} .a\operatorname {Sin} .d}},\\\\2\operatorname {Cos} .^{2}{\frac {1}{2}}(a,d)&=1+\operatorname {Cos} .(a,d)&={\frac {\operatorname {Cos} .x-\operatorname {Cos} .(a+d)}{\operatorname {Sin} .a\operatorname {Sin} .d}}\,;\end{array}}$

mais, on a généralement

$\operatorname {Cos} .k=-2\operatorname {Sin} .^{2}{\frac {1}{2}}k\,;$

donc

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

En mettant, dans ces deux formules, pour $\operatorname {Sin} .^{2}{\frac {1}{2}}x,$ la valeur (I) que nous avons trouvée tout à l’heure, elles deviennent, en rassemblant d’une part tous les termes multipliés par $\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c,$ et de l’autre tous les termes multipliés par $\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d,$

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

mais on trouve facilement