Cela posé, on sait, comme nous l’avons déjà observé (pag. 41), que
étant un angle quelconque et
un nombre entier positif quelconque, on a
![{\displaystyle \operatorname {Cos} .^{p}z={\frac {1}{2^{p-1}}}\left\{\operatorname {Cos} .pz+{\frac {p}{1}}\operatorname {Cos} .(p-2)z+{\frac {p}{1}}.{\frac {p-1}{2}}.\operatorname {Cos} .(p-4)z+\ldots \right\},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a48611a52a482554ee6bbe7c58179e50e7279544)
pourvu qu’on arrête le développement dès qu’on ne rencontrera plus d’arcs positifs, et que, lorsque
sera pair, on ne prenne que la moitié du terme qui contiendra l’arc nul.
Mettant successivement pour
dans cette formule, les termes de la progression, progression,
on aura
![{\displaystyle \operatorname {Cos} .^{p}a={\frac {1}{2^{p-1}}}\left\{\operatorname {Cos} .pa+{\frac {p}{1}}\operatorname {Cos} .(p-2)a+{\frac {p}{1}}.{\frac {p-1}{2}}.\operatorname {Cos} .(p-4)a+\ldots \right\},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c01cb1092153f719af5f4ed7615a4458a58b0bc9)
![{\displaystyle \operatorname {Cos} .^{p}(a+x)={\frac {1}{2^{p-1}}}\left\{\operatorname {Cos} .p(a+x)+{\frac {p}{1}}\operatorname {Cos} .(p-2)(a+x)+\ldots \right\},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/deec54e9753bfc96bd3b7b0e5f0953385a8017b2)
![{\displaystyle \operatorname {Cos} .^{p}(a+2x)={\frac {1}{2^{p-1}}}\left\{\operatorname {Cos} .p(a+2x)+{\frac {p}{1}}\operatorname {Cos} .(p-2)(a+2x)+\ldots \right\}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5a8985b8ea827aa96f5155064cb2a90d4c23fdbf)
![{\displaystyle \operatorname {Cos} .^{p}(a+{\overline {n-1}}x)=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/960dcd524ec1000359d931d1456006c977f290c3)
![{\displaystyle {\frac {1}{2^{p-1}}}\left\{\operatorname {Cos} .p(a+{\overline {n-1}}x)+{\frac {p}{1}}\operatorname {Cos} .(p-2)(a+{\overline {n-1}}x)+\ldots \right\}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/911c9e4f88edf460ad40b645bb474b4692d97d36)
ce qui donnera, en ajoutant,
![{\displaystyle \operatorname {Cos} .^{p}a+\operatorname {Cos} .^{p}(a+x)+\operatorname {Cos} .^{p}(a+2x)+\ldots +\operatorname {Cos} .^{p}(a+{\overline {n-1}}.x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e8408766dad70a8d3a9d926cf9427bcf5d16942a)
![{\displaystyle ={\frac {1}{2^{p-1}}}\left\{{\begin{array}{l}&\quad \operatorname {Cos} .pa+\operatorname {Cos} .p(a+x)+\operatorname {Cos} .p(a+2x)+\\&\qquad \qquad \qquad \qquad \ldots +\operatorname {Cos} .p(a+{\overline {n-1}}.x)\\&+{\frac {p}{1}}\left\{\operatorname {Cos} .(p-2)a+\operatorname {Cos} .(p-2)(a+x)+\right.\\&\qquad \qquad \qquad \qquad \left.\ldots +\operatorname {Cos} .(p-2)(a+{\overline {n-1}}.x)\right\}\\&+{\frac {p}{1}}.{\frac {p-1}{2}}\left\{\operatorname {Cos} .(p-4)a+\operatorname {Cos} .(p-4)(a+x)+\right.\\&\qquad \qquad \qquad \qquad \left.\ldots +\operatorname {Cos} .(p-4)(a+{\overline {n-1}}.x)\right\}\\&+\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \end{array}}\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/890967a07f1f5b16437a51b970f5e00def4bb526)