Page:Joseph Louis de Lagrange - Œuvres, Tome 4.djvu/291

et de là (7)

${\begin{aligned}&{\frac {\gamma -\alpha -\beta }{2}}+90^{\circ }\\&={\frac {\operatorname {tang} {\cfrac {c}{2}}}{\operatorname {tang} {\cfrac {b}{2}}}}\sin \alpha +{\frac {1}{2}}\left({\frac {\operatorname {tang} {\cfrac {c}{2}}}{\operatorname {tang} {\cfrac {b}{2}}}}\right)^{2}\sin 2\alpha +{\frac {1}{3}}\left({\frac {\operatorname {tang} {\cfrac {c}{2}}}{\operatorname {tang} {\cfrac {b}{2}}}}\right)^{3}\sin 3\alpha +\ldots \\&={\frac {\operatorname {tang} {\cfrac {c}{2}}}{\operatorname {tang} {\cfrac {b}{2}}}}\sin(\beta -\gamma )+{\frac {1}{2}}\left({\frac {\operatorname {tang} {\cfrac {c}{2}}}{\operatorname {tang} {\cfrac {b}{2}}}}\right)^{2}\sin 2(\beta -\gamma )\\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad +{\frac {1}{3}}\left({\frac {\operatorname {tang} {\cfrac {c}{2}}}{\operatorname {tang} {\cfrac {b}{2}}}}\right)^{3}\sin 3(\beta -\gamma )+\ldots .\end{aligned}}$

22. La seconde étant comparée de même à celle du n^o 6, on aura les mêmes valeurs de $\varphi ,\omega ,y$ que ci-dessus, mais celle de $x$ sera ${\frac {\beta +\gamma }{2}}\,;$ on aura donc dans ce cas

$\theta =-\operatorname {tang} {\frac {c}{2}}\operatorname {tang} {\frac {b}{2}}\,;$

donc (7)

${\begin{aligned}&{\frac {\gamma +\alpha +\beta }{2}}-90^{\circ }\\&\quad =\operatorname {tang} {\frac {b}{2}}\operatorname {tang} {\frac {c}{2}}\sin \alpha -{\frac {1}{2}}\left(\operatorname {tang} {\frac {b}{2}}\operatorname {tang} {\frac {c}{2}}\right)^{2}\sin 2\alpha \\&\qquad +{\frac {1}{3}}\left(\operatorname {tang} {\frac {b}{2}}\operatorname {tang} {\frac {c}{2}}\right)^{3}\sin 3\alpha -\ldots .\\\\&\quad =\operatorname {tang} {\frac {b}{2}}\operatorname {tang} {\frac {c}{2}}\sin(\beta +\gamma )-{\frac {1}{2}}\left(\operatorname {tang} {\frac {b}{2}}\operatorname {tang} {\frac {c}{2}}\right)^{2}\sin 2(\beta +\gamma )\\&\qquad +{\frac {1}{3}}\left(\operatorname {tang} {\frac {b}{2}}\operatorname {tang} {\frac {c}{2}}\right)^{3}\sin 3(\beta +\gamma )-\ldots .\end{aligned}}$

23. La troisième et la quatrième équation donneront des formules analogues aux précédentes en y changeant seulement $\beta$ en $b,$ $\gamma$ en $c$ et $\alpha$ en $180^{\circ }-a.$ Ainsi l’on aura, en premier lieu,

${\begin{aligned}&{\frac {c+a-b}{2}}\\&={\frac {\operatorname {tang} {\cfrac {\gamma }{2}}}{\operatorname {tang} {\cfrac {\beta }{2}}}}\sin a-{\frac {1}{2}}\left({\frac {\operatorname {tang} {\cfrac {\gamma }{2}}}{\operatorname {tang} {\cfrac {\beta }{2}}}}\right)^{2}\sin 2a+{\frac {1}{3}}\left({\frac {\operatorname {tang} {\cfrac {\gamma }{2}}}{\operatorname {tang} {\cfrac {\beta }{2}}}}\right)^{3}\sin 3a-\ldots \\&={\frac {\operatorname {tang} {\cfrac {\gamma }{2}}}{\operatorname {tang} {\cfrac {\beta }{2}}}}\sin(b-c)+{\frac {1}{2}}\left({\frac {\operatorname {tang} {\cfrac {\gamma }{2}}}{\operatorname {tang} {\cfrac {\beta }{2}}}}\right)^{2}\sin 2(b-c)\\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad +{\frac {1}{3}}\left({\frac {\operatorname {tang} {\cfrac {\gamma }{2}}}{\operatorname {tang} {\cfrac {\beta }{2}}}}\right)^{3}\sin 3(b-c)+\ldots .\end{aligned}}$