Page:Joseph Louis de Lagrange - Œuvres, Tome 2.djvu/27

on aura de plus

${\begin{aligned}\cos n(x+y)\,\ =&{\frac {1+u^{2}v^{2}}{2uv}},\\\sin n(x+y)\,\ =&{\frac {1-u^{2}v^{2}}{2uv}}{\sqrt {-1}},\\\cos {\frac {n(x-y)}{2}}=&{\frac {v+u}{2{\sqrt {vu}}}},\\\sin {\frac {n(x-y)}{2}}=&{\frac {v-u}{2{\sqrt {vu}}}}{\sqrt {-1}}.\end{aligned}}$

Enfin on aura

$dx={\frac {du}{nu{\sqrt {-1}}}},\qquad dy={\frac {dv}{nv{\sqrt {-1}}}}.$

Supposons outre cela

${\begin{alignedat}{2}\cos(\beta +\alpha )=&\mathrm {A} ,&\cos 2(\gamma +\alpha )=&\mathrm {B} ,\\\cos(\beta -\alpha )=&\mathrm {E} ,\qquad &\cos 2(\gamma -\alpha )=&\mathrm {F} ,\end{alignedat}}$

on aura

${\begin{aligned}\cos 2\alpha =&\mathrm {AE+{\sqrt {1-A^{2}}}{\sqrt {1-E^{2}}}} ,\\\cos 4\alpha =&\mathrm {BF+{\sqrt {1-B^{2}}}{\sqrt {1-F^{2}}}} \,;\end{aligned}}$

donc

(G)

$1+\mathrm {BF+{\sqrt {1-B^{2}}}{\sqrt {1-F^{2}}}=2(AE+{\sqrt {1-A^{2}}}{\sqrt {1-E^{2}}}} .$

Ensuite, en faisant, pour abréger,

${\begin{aligned}&\mathrm {AE+{\sqrt {1-A^{2}}}{\sqrt {1-E^{2}}}=M} ,\\&\mathrm {BF+{\sqrt {1-B^{2}}}{\sqrt {1-F^{2}}}=N} ,\end{aligned}}$

on aura

${\begin{alignedat}{2}\cos \alpha \,\ =&{\sqrt {\frac {1+M}{2}}},&\sin \alpha \,\ =&{\sqrt {\frac {1-M}{2}}},\\\cos 2\gamma =&{\sqrt {\frac {1+N}{2}}},\qquad &\sin 2\gamma =&{\sqrt {\frac {1-N}{2}}},\end{alignedat}}$

donc, en faisant toutes ces substitutions dans les formules du numéro précédent, on aura d’abord

${\begin{aligned}\mathrm {X} =&a+b{\frac {\mathrm {A} \left(1+u^{2}\right)-{\sqrt {\mathrm {A} ^{2}-1}}\left(1-u^{2}\right)}{2u}}+c{\frac {\mathrm {B} \left(1+u^{4}\right)-{\sqrt {\mathrm {B} ^{2}-1}}\left(1-u^{4}\right)}{2u^{2}}},\\\mathrm {Y} =&a+b{\frac {\mathrm {E} \left(1+v^{2}\right)-{\sqrt {\mathrm {E} ^{2}-1}}\left(1-v^{2}\right)}{2v}}+c{\frac {\mathrm {F} \left(1+v^{4}\right)-{\sqrt {\mathrm {F} ^{2}-1}}\left(1-v^{4}\right)}{2v^{2}}},\end{aligned}}$