24
INTÉGRATION
![{\displaystyle {\begin{aligned}M&=\ 1\ -{\frac {x^{2}}{2}}+{\frac {x^{4}}{4!}}-{\frac {x^{6}}{6!}}+{\frac {x^{8}}{8!}}-\ldots ,\\N&={\frac {x}{1}}-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+{\frac {x^{9}}{9!}}-\ldots ,\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b9cc7bb212e34271b4b40255f7a089a71467d138)
on doit avoir
![{\displaystyle \operatorname {Sin} .x={\frac {N}{M^{2}+N^{2}}},\qquad \operatorname {Cos} .x={\frac {M}{M^{2}+N^{2}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9ebb57e42b0bf02bffc2049623a463a873ce2218)
d’où
![{\displaystyle \operatorname {Tang} .x={\frac {N}{M}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eb71ee752617252cf5e65222945ec80c1e742f42)
or, s’il en est ainsi, on devra avoir
![{\displaystyle 1=\operatorname {Sin} .^{2}x+\operatorname {Cos} .^{2}x=\left({\frac {N}{M^{2}+N^{2}}}\right)^{2}+\left({\frac {M}{M^{2}+N^{2}}}\right)^{2}={\frac {1}{M^{2}+N^{2}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/52e01bcf86ef52b0c708aa4c95646bf8adff5abc)
d’où on conclura
![{\displaystyle M^{2}+N^{2}=1\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4a6790708bda44b594e5f7b422738b35bc6e7cdb)
on aura donc simplement
![{\displaystyle \operatorname {Sin} .x=N,\qquad \operatorname {Cos} .x=M\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/21ce76564696b45d24420fe352d5cf593ac9db44)
c’est-à-dire
![{\displaystyle {\begin{aligned}\operatorname {Sin} .x&={\frac {x}{1}}-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+\ldots \\\operatorname {Cos} .x&=\ 1\ -{\frac {x^{2}}{2}}+{\frac {x^{4}}{4!}}-{\frac {x^{6}}{6!}}-\ldots \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/027fdf873ee1fd6f453767d2eeec4c98957593c5)
ce qui, en effet, est rigoureusement vrai
PROBLÈME IV. Déterminer la longueur d’un arc de cercle dont la tangente est donnée ?