Page:Annales de mathématiques pures et appliquées, 1811-1812, Tome 2.djvu/84

78

CENTRE DES MOYENNES DISTANCES

${\begin{aligned}2r^{2}.\operatorname {Sin} .^{4}{\tfrac {1}{2}}\mathrm {AM} &\cdot {\frac {\mathrm {MM} '.\operatorname {Sin} .\mathrm {M} }{\operatorname {Sin} .\mathrm {AM} }}\\\\&=2r^{2}.\operatorname {Sin} .^{4}{\tfrac {1}{2}}\mathrm {AM} \cdot {\frac {\operatorname {Sin} .\mathrm {AM} .\operatorname {Sin} .\mathrm {M} }{\operatorname {Sin} .^{2}\mathrm {AM} }}.\mathrm {MM} '\\\\&=2r^{2}.\operatorname {Sin} .^{4}{\tfrac {1}{2}}\mathrm {AM} \cdot {\frac {\operatorname {Sin} .\mathrm {B} .\operatorname {Sin} .\mathrm {AB} }{\operatorname {Sin} .^{2}\mathrm {AM} }}.\mathrm {MM} '\\\\&={\tfrac {1}{2}}r^{2}.\mathrm {MM} '.\operatorname {Sin} .\mathrm {B} .\operatorname {Sin} .\mathrm {AB} .\operatorname {Tang} ^{2}{\tfrac {1}{2}}\mathrm {AM} \\\\&={\tfrac {1}{2}}r^{2}.\operatorname {Sin} .\mathrm {B} .\operatorname {Sin} .\mathrm {AB} .{\frac {1-\operatorname {Cos} .\mathrm {AM} }{1+\operatorname {Cos} .\mathrm {AM} }}.\mathrm {MM} ',\\\\&={\tfrac {1}{2}}r^{2}.\mathrm {MM} '.\operatorname {Sin} .\mathrm {B} .\operatorname {Sin} .\mathrm {AB} .\left\{{\frac {2}{1+\operatorname {Cos} .\mathrm {AM} }}-1\right\}\end{aligned}}$

$={\frac {r^{2}\operatorname {Sin} .\mathrm {B} .\operatorname {Sin} .\mathrm {AB} .\mathrm {MM} '}{1+\operatorname {Sin} .\mathrm {AB} .\operatorname {Sin} .\mathrm {BM} .\operatorname {Cos} .\mathrm {B} +\operatorname {Cos} .\mathrm {AB} .\operatorname {Cos} .\mathrm {BM} }}-{\tfrac {1}{2}}r^{2}.\mathrm {MM} '\operatorname {Sin} .\mathrm {B} .\operatorname {Sin} .\mathrm {AB} ~;$

en observant donc que

$1-\operatorname {Sin} .^{2}\mathrm {AB} .\operatorname {Cos} .^{2}\mathrm {B} -\operatorname {Cos} .^{2}\mathrm {AB} =\operatorname {Sin} .^{2}\mathrm {AB} .\operatorname {Sin} .^{2}\mathrm {B} ,$

on trouvera pour le moment du triangle $\mathrm {ABM}$ (§. 2.)

$2r^{3}.\operatorname {Arc} \left\{\operatorname {Tang} .={\frac {\operatorname {Sin} .\mathrm {AB} .\operatorname {Sin} .\mathrm {B} .\operatorname {Tang} .{\tfrac {1}{2}}\mathrm {BM} }{1+\operatorname {Cos} .\mathrm {AB} +\operatorname {Cos} .\mathrm {B} .\operatorname {Sin} .\mathrm {AB} .\operatorname {Tang} .{\tfrac {1}{2}}\mathrm {BM} }}\right\}$

$-{\frac {1}{2}}r^{2}.\operatorname {Sin} .\mathrm {B} .\operatorname {Sin} .\mathrm {AB} .\mathrm {BM} .$

Le moment du triangle $\mathrm {ABB} '$, relativement au même plan, sera donc,

$2r^{3}.\operatorname {Arc} \left\{\operatorname {Tang} .={\frac {\operatorname {Sin} .\mathrm {AB} .\operatorname {Sin} .\mathrm {B} .\operatorname {Tang} .{\tfrac {1}{2}}\mathrm {BB} '}{1+\operatorname {Cos} .\mathrm {AB} +\operatorname {Cos} .\mathrm {B} .\operatorname {Sin} .\mathrm {AB} .\operatorname {Tang} .{\tfrac {1}{2}}\mathrm {BB} '}}\right\}$

$-{\frac {1}{2}}r^{2}.\operatorname {Sin} .\mathrm {B} .\operatorname {Sin} .\mathrm {AB} .\mathrm {BB} '.$

§. 5.

Comme on a $\operatorname {Sin} .\mathrm {B} .\operatorname {Sin} .\mathrm {AB} =\operatorname {Sin} .\mathrm {B} '.\operatorname {Sin} .\mathrm {A} '\mathrm {B} ',$ tout est symétrique, dans cette expression par rapport aux angles $\mathrm {B} ,\mathrm {B} ',$ et aux côtés opposés $\mathrm {AB} ',\mathrm {AB} ,$ excepté le dénominateur ; mais nous allons faire voir que ce dénominateur peut aussi être rendu symétrique, ainsi que cela doit être.

En effet, $\operatorname {Cos} .\mathrm {B} ={\tfrac {\operatorname {Cos} .\mathrm {AB} '-\operatorname {Cos} .\mathrm {AB} .\operatorname {Cos} .\mathrm {BB} '}{\operatorname {Sin} .\mathrm {AB} .\operatorname {Sin} .\mathrm {BB} '}}~;$

donc $\operatorname {Cos} .\mathrm {B} .\operatorname {Sin} .\mathrm {AB} ={\tfrac {\operatorname {Cos} .\mathrm {AB} '-\operatorname {Cos} .\mathrm {AB} .\operatorname {Cos} .\mathrm {BB} '}{\operatorname {Sin} .\mathrm {BB} '}}={\tfrac {\operatorname {Cos} .\mathrm {AB} '-\operatorname {Cos} .\mathrm {AB} .\operatorname {Cos} .\mathrm {BB} '}{2\operatorname {Sin} .{\tfrac {1}{2}}\mathrm {BB} '.\operatorname {Cos} .{\tfrac {1}{2}}\mathrm {BB} '}}~;$