La formule (36) donne
![{\displaystyle \operatorname {d} .\operatorname {Sin} .X'=\operatorname {d} X'.\operatorname {Cos} .X'\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2ca0208bf46137e6aec60a10ff2ac43058399cf4)
posant
![{\displaystyle \operatorname {Sin} .X'=X,\quad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/80f797d2a11491fcacb2280c6b377aafc6e38f07)
d’où
![{\displaystyle \quad X'=\operatorname {Ang} .(\operatorname {Sin} .=X),\quad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/0d2297f3cc5e3b06ca3703eebfaca1dfbd01fd80)
et
![{\displaystyle \quad \operatorname {Cos} .X'={\sqrt {1-X^{2}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/98f5491f4594578b79f02fa126bfda26583e498b)
il viendra, en substituant,
![{\displaystyle \operatorname {d} X={\sqrt {1-X^{2}}}.\operatorname {d} .\operatorname {Ang} .(\operatorname {Sin} .=X)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a77fe254cb25bbca0115e54933a404b075ffd88)
ce qui donnera
on aurait pu partir aussi des définitions connues
![{\displaystyle \operatorname {Sin} .X={\frac {e^{X{\sqrt {-1}}}-e^{-X{\sqrt {-1}}}}{2{\sqrt {-1}}}},\qquad \operatorname {Cos} .X={\frac {e^{X{\sqrt {-1}}}+e^{-X{\sqrt {-1}}}}{2}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a79d8391dff36702eb30fa73b0119ed408c63daa)
qui auraient conduit aux mêmes résultats. En considérant ensuite que
![{\displaystyle \operatorname {Tang} .X={\frac {\operatorname {Sin} .X}{\operatorname {Cos} .X}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f070d0f0b2762deb6260b3ad5d0873874f986f2a)
on en aurait conclu (19)
![{\displaystyle \operatorname {d} .\operatorname {Tang} .X={\frac {\operatorname {Cos} .X.\operatorname {d} .\operatorname {Sin} .X-\operatorname {Sin} .X.\operatorname {d} .\operatorname {Cos} .X}{\operatorname {Cos} .^{2}X}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ac30347b13d85b79aeefa643194df7bd5ff30fc5)
ou bien (36), (37)
![{\displaystyle \operatorname {d} .\operatorname {Tang} .X={\frac {\left(\operatorname {Cos} .^{2}X+\operatorname {Sin} .^{2}X\right)\operatorname {d} .X}{\operatorname {Cos} .^{2}X}}={\frac {\operatorname {d} .X}{\operatorname {Cos} .^{2}X}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c7d49c8a8cb57a5d0f1e86f72c0859bfce92f0c0)
comme nous l’avons déjà trouvé ci-dessus (34).