Page:Annales de mathématiques pures et appliquées, 1828-1829, Tome 19.djvu/373

${\sqrt {4m\operatorname {Cos} .\alpha -x^{2}}}={\sqrt {2m\operatorname {Cos} .\alpha (1+\operatorname {Cos} .\omega )}}=2\operatorname {Cos} .{\frac {1}{2}}\omega {\sqrt {m\operatorname {Cos} .\alpha }}\,;$

en conséquence l’équation (12) deviendra

$2(1+B){\sqrt {m\operatorname {Cos} .\alpha }}=\operatorname {Log} .{\frac {\operatorname {Cos} .{\frac {1}{2}}\omega -1}{\operatorname {Cos} .{\frac {1}{2}}\omega +1}}=\operatorname {Log} .{\frac {\operatorname {Cos} .{\frac {1}{2}}\left(\theta -{\frac {2\varpi t\operatorname {Sin} .\alpha }{T}}\right)-1}{\operatorname {Cos} .{\frac {1}{2}}\left(\theta -{\frac {2\varpi t\operatorname {Sin} .\alpha }{T}}\right)+1}},$

Pour faire disparaître la constante $B$ rappelons-nous qu’on doit avoir, en même temps, $t=0$ et $\theta =\beta ,$ ce qui donne, en substituant,

$2B{\sqrt {m\operatorname {Cos} .\alpha }}=\operatorname {Log} .{\frac {\operatorname {Cos} .{\frac {1}{2}}\beta -1}{\operatorname {Cos} .{\frac {1}{2}}\beta +1}}\,;$

retranchant celle équation de la précédente, on aura

$2t{\sqrt {m\operatorname {Cos} .\alpha }}=\operatorname {Log} .\left\{{\frac {1+\operatorname {Cos} .{\frac {1}{2}}\beta }{1-\operatorname {Cos} .{\frac {1}{2}}\beta }}.{\frac {1-\operatorname {Cos} .{\frac {1}{2}}\left(\theta -{\frac {2\varpi t\operatorname {Sin} .\alpha }{T}}\right)}{1+\operatorname {Cos} .{\frac {1}{2}}\left(\theta -{\frac {2\varpi t\operatorname {Sin} .\alpha }{T}}\right)}}\right\}\,;$

ou bien

$2t{\sqrt {m\operatorname {Cos} .\alpha }}=\operatorname {Log} .{\frac {\operatorname {Tang} .^{2}{\frac {1}{4}}\left(\theta -{\frac {2\varpi t\operatorname {Sin} .\alpha }{T}}\right)}{\operatorname {Tang} .^{2}{\frac {1}{4}}\beta }}$

$=2\operatorname {Log} .{\frac {\operatorname {Tang} .{\frac {1}{4}}\left(\theta -{\frac {2\varpi t\operatorname {Sin} .\alpha }{T}}\right)}{\operatorname {Tang} .{\frac {1}{4}}\beta }}\,;$

c’est-à-dire,

$t{\sqrt {m\operatorname {Cos} .\alpha }}=\operatorname {Log} .{\frac {\operatorname {Tang} .{\frac {1}{4}}\left(\theta -{\frac {2\varpi t\operatorname {Sin} .\alpha }{T}}\right)}{\operatorname {Tang} .{\frac {1}{4}}\beta }},$

ce qui donne

$\operatorname {Tang} .{\frac {1}{4}}\left(\theta -{\frac {2\varpi t\operatorname {Sin} .\alpha }{T}}\right)=e^{t{\sqrt {m\operatorname {Cos} .\alpha }}}.\operatorname {Tang} .{\frac {1}{4}}\beta ,$

et, par suite,