Page:Lagrange - Œuvres (1867) vol. 1.djvu/604

donc

$\sin \varphi =\varphi -\alpha \sin ^{3}\varphi -\beta \sin ^{5}\varphi -\gamma \sin ^{7}\varphi -\ldots \,;$

donc, supposant $\varphi ={\frac {m\pi }{2(n+1)}},$ et faisant, pour abréger,

$\sin {\frac {m\pi }{2(n+1)}}=x,$

on aura

$\sin {\frac {m\pi }{2(n+1)}}={\frac {m\pi }{2(n+1)}}-\alpha x^{3}-\beta x^{5}-\gamma x^{7}-\ldots ,$

et, par conséquent,

${\begin{aligned}\cos \left[2tk\sin {\frac {m\pi }{2(n+1)}}\right]&=\cos \left({\frac {mkt}{n+1}}\pi -2\alpha ktx^{3}-2\beta ktx^{5}-2\gamma ktx^{7}-\ldots \right)\\&=\cos \left({\frac {mkt}{n+1}}\pi \right)\cos \left(2\alpha ktx^{3}+2\beta ktx^{5}+\ldots \right)\\&=\sin \left({\frac {mkt}{n+1}}\pi \right)\sin \left(2\alpha ktx^{3}+2\beta ktx^{5}+\ldots \right).\end{aligned}}$

Or

$\cos \left(2\alpha ktx^{3}+2\beta ktx^{5}+\ldots \right)=1-2\alpha ^{2}k^{2}t^{2}x^{6}-2\alpha \beta k^{2}t^{2}x^{8}-2\beta ^{2}k^{2}t^{2}x^{10}-\ldots ,$

et

$\sin \left(2\alpha ktx^{3}+2\beta ktx^{5}+\ldots \right)$

$=2\alpha ktx^{3}+2\beta ktx^{5}+2\gamma ktx^{7}+\left(2\delta kt-{\frac {4\alpha ^{3}k^{3}t^{3}}{3}}\right)x^{9}+\ldots \,;$

de plus

${\frac {1}{\sqrt {1-x^{2}}}}=1+{\frac {1}{2}}x^{2}+{\frac {1.3}{2.4}}x^{4}+{\frac {1.3.5}{2.4.6}}x^{6}+\ldots ,$

et, par conséquent,

$\left(1+{\frac {1}{2}}x^{2}+{\frac {1.3}{2.4}}x^{4}+\ldots \right){\sqrt {1-x^{2}}}\,;$

donc, on aura aussi

$\sin \left(2\alpha ktx^{3}+2\beta ktx^{5}+\ldots \right)$

${\begin{aligned}&=\left\{2\alpha ktx^{2}+(\alpha +2\beta )ktx^{4}+\left({\frac {3\alpha }{4}}+\beta +2\gamma \right)ktx^{6}\right.\\&+\left.\left[\left({\frac {3.5\alpha }{4.6}}+{\frac {3\beta }{4}}+\gamma +2\delta \right)kt-{\frac {4\alpha ^{3}}{3}}k^{3}t^{3}\right]x^{8}+\ldots \right\}x{\sqrt {1-x^{2}}}\end{aligned}}$