QUESTIONS RÉSOLUES.
Solution des trois problèmes d’analise transcendante
énoncés à la page 247 du présent volume ;
M. Querret, chef d’institution, à St-Malo.
≈≈≈≈≈≈≈≈≈
PROBLÈME. Assigner la somme finie de chacune des trois suites infinies que voici :
1.
o ![{\displaystyle {\frac {a\operatorname {Cos} .x}{1}}-{\frac {a^{3}\operatorname {Cos} .3x}{3}}+{\frac {a^{5}\operatorname {Cos} .5x}{5}}-{\frac {a^{7}\operatorname {Cos} .7x}{7}}+\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/54d768ca7c660aa3b11cfda41c482996a278f257)
2.
o ![{\displaystyle \operatorname {Cos} .x+{\frac {1}{2}}{\frac {\operatorname {Cos} .3x}{3}}+{\frac {1.3}{2.4}}.{\frac {\operatorname {Cos} .5x}{5}}+{\frac {1.3.5}{2.4.6}}.{\frac {\operatorname {Cos} .7x}{7}}+\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/c0f4470a8e4a9289be818836fd568b455acfb078)
3.
o ![{\displaystyle {\frac {\operatorname {Cos} .x\operatorname {Cos} .y}{1}}-{\frac {\operatorname {Cos} .2x\operatorname {Cos} .2y}{2}}+{\frac {\operatorname {Cos} .3x\operatorname {Cos} .3y}{3}}-{\frac {\operatorname {Cos} .4x\operatorname {Cos} .4y}{4}}+\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/6b905d0351a76d3c4950cad7ff173e897bcdf17e)
Solution. Nous allons déduire la sommation de chacune de ces trois suites du théorème que nous avons établi à la page 107 de ce volume, et que nous rappelons en ces termes :
Si l’on représente par
la somme de la suite infinie
![{\displaystyle A_{0}+A_{1}a+A_{2}a^{2}+A_{3}a^{3}+A_{4}a^{4}+\ldots ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/57c16051bbe5528255aa42666cd3e4b74f77e7c0)
dans laquelle
sont supposés représenter des coefficiens numériques ; les sommes des deux séries
![{\displaystyle A_{0}+A_{1}a\operatorname {Cos} .x+A_{2}a^{2}\operatorname {Cos} .2x+A_{3}a^{3}\operatorname {Cos} .3x+A_{4}a^{4}\operatorname {Cos} .4x+\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/a85231632c86244ca75d861e81bfe3661a9ece11)
![{\displaystyle A_{0}+A_{1}a\operatorname {Sin} .x\ +A_{2}a^{2}\operatorname {Sin} .2x\ +A_{3}a^{3}\operatorname {Sin} .3x\ +A_{4}a^{4}\operatorname {Sin} .4x\ +\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/aa295205183aea53dbd9ab6b9257874b6717053b)
seront respectivement
![{\displaystyle {\frac {\operatorname {f} \left[a\left(\operatorname {Cos} .x+{\sqrt {-1}}\operatorname {Sin} .x\right)\right]+\operatorname {f} \left[a\left(\operatorname {Cos} .x-{\sqrt {-1}}\operatorname {Sin} .x\right)\right]}{2}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e6d87c780e46e984dcf0dd464602358cc0f0c3cb)
et
![{\displaystyle {\frac {\operatorname {f} \left[a\left(\operatorname {Cos} .x+{\sqrt {-1}}\operatorname {Sin} .x\right)\right]-\operatorname {f} \left[a\left(\operatorname {Cos} .x-{\sqrt {-1}}\operatorname {Sin} .x\right)\right]}{2{\sqrt {-1}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8c5bfdb18f06c3ef5ddc67050d7b3ab009609bb4)
Cela posé, on a
![{\displaystyle {\frac {a}{1}}-{\frac {a^{3}}{3}}+{\frac {a^{5}}{5}}-{\frac {a^{7}}{7}}+{\frac {a^{9}}{9}}-\ldots =\operatorname {Arc} .(\operatorname {Tang} .=a),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/597843e8936752b78de45748d16f57929b387536)
donc 1.o la somme de la série
![{\displaystyle {\frac {a\operatorname {Cos} .x}{1}}-{\frac {a^{3}\operatorname {Cos} .3x}{3}}+{\frac {a^{5}\operatorname {Cos} .5x}{5}}-{\frac {a^{7}\operatorname {Cos} .7x}{7}}+\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/54d768ca7c660aa3b11cfda41c482996a278f257)
sera
![{\displaystyle {\frac {\operatorname {Arc} .\left[\operatorname {Tang} .=a\left(\operatorname {Cos} .x+{\sqrt {-1}}\operatorname {Sin} .x\right)\right]+\operatorname {Arc} .\left[\operatorname {Tang} .=a\left(\operatorname {Cos} .x-{\sqrt {-1}}\operatorname {Sin} .x\right)\right]}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a6dbe989f9a2bcc889e580c23b86741096652964)
Soient
le premier de ces arcs et
le second, on aura pour la somme de la série
et de plus
![{\displaystyle \operatorname {Tang} .M=a\left(\operatorname {Cos} .x+{\sqrt {-1}}\operatorname {Sin} .x\right),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d3cb8ccde95c84fadce6c460b01e5698a9922548)
![{\displaystyle \operatorname {Tang} .N=a\left(\operatorname {Cos} .x-{\sqrt {-1}}\operatorname {Sin} .x\right)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b2e4914fcb258d5a672272789c598f59d20a2399)
d’où
![{\displaystyle \operatorname {Tang} .M+\operatorname {Tang} .N=2a\operatorname {Cos} .x,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5dd329914af9ea9caab5917a13a26654f86acec2)
d’où encore
![{\displaystyle 1-\operatorname {Tang} .M\operatorname {Tang} .N=1-a^{2}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/11a19fa484768470677838292314137cbf7a41de)
donc
![{\displaystyle \operatorname {Tang} .(M+N)={\frac {\operatorname {Tang} .M+\operatorname {Tang} .N}{1-\operatorname {Tang} .M\operatorname {Tang} .N}}={\frac {2a\operatorname {Cos} .x}{1-a^{2}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2082cbcc1d99c24fb6990b8d2e1787bc7c98160f)
donc
![{\displaystyle M+N=\operatorname {Arc} .\left(\operatorname {Tang} .={\frac {2a\operatorname {Cos} .x}{1-a^{2}}}\right)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6b2df4be004ae645cd800c157f34250715f3c010)
donc enfin
![{\displaystyle {\frac {M+N}{2}}={\frac {1}{2}}\operatorname {Arc} .\left(\operatorname {Tang} .={\frac {2a\operatorname {Cos} .x}{1-a^{2}}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3f3b0c858b2bb0087d3898186f88935c261bea03)
qui sera conséquemment la somme finie demandée de la première des trois séries infinies proposées.
Si l’on suppose
on a
![{\displaystyle {\frac {M+N}{2}}={\frac {1}{2}}\operatorname {Arc} .\left(\operatorname {Tang} .=\infty \right)={\frac {1}{4}}\varpi \,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/23c15f83394190b2e7209bcf331a380c96e29722)
donc, comme on le savait déjà,
![{\displaystyle {\frac {\varpi }{4}}=\operatorname {Cos} .x-{\frac {\operatorname {Cos} .3x}{3}}+{\frac {\operatorname {Cos} .5x}{5}}-{\frac {\operatorname {Cos} .7x}{7}}+\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/2b05108beca75f5a11e2a9d0d835ad9c45e83489)
quel que soit ![{\displaystyle x.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d07e9f568a88785ae48006ac3c4b951020f1699a)
En second lieu
![{\displaystyle a+{\frac {1}{2}}.{\frac {a^{3}}{3}}+{\frac {1.3}{2.4}}.{\frac {a^{5}}{5}}+{\frac {1.3.5}{2.4.6}}.{\frac {a^{7}}{7}}+\ldots =\operatorname {Arc} .(\operatorname {Sin} .=a)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4c0fed309cae9099d1d4725a44d659df4e4b10ec)
ainsi qu’il est aisé de s’en assurer, en intégrant par les séries la fonction différentielle
![{\displaystyle {\frac {\operatorname {d} a}{\sqrt {1-a^{2}}}}=\operatorname {d} .\operatorname {Arc} .(\operatorname {Sin} .=a)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d93e130e6400600ea95d2e6f54347fd27a1320f6)
donc, la, somme de la série
![{\displaystyle \operatorname {Cos} .x+{\frac {1}{2}}.{\frac {\operatorname {Cos} .3x}{3}}+{\frac {1.3}{2.4}}.{\frac {\operatorname {Cos} .5x}{5}}+{\frac {1.3.5}{2.4.6}}.{\frac {\operatorname {Cos} .7x}{7}}+\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/17e8e5f24743d2cb1388cafe56ba9164b382c282)
sera
![{\displaystyle {\frac {\operatorname {Arc} .\left(\operatorname {Sin} .=\operatorname {Cos} .x+{\sqrt {-1}}\operatorname {Sin} .x\right)+\operatorname {Arc} .\left(\operatorname {Sin} .=\operatorname {Cos} .x-{\sqrt {-1}}\operatorname {Sin} .x\right)}{2}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aaed86d8a7c2a3d74e2386bc80539a233db6ee61)
Soient
le premier de ces arcs et
le second ; la somme cherchée sera donc
![{\displaystyle {\frac {P+Q}{2}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d8254b295076b4fc7ff9864b16222b7d7381fe69)
et l’on aura
![{\displaystyle \operatorname {Sin} .P=\operatorname {Cos} .x+{\sqrt {-1}}\operatorname {Sin} .x,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/884305b65e8a29aedb485c34d2655271f31b87ee)
![{\displaystyle \operatorname {Sin} .Q=\operatorname {Cos} .x-{\sqrt {-1}}\operatorname {Sin} .x,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/64cdb8a9e53774d630a2ab9c8cdd2e1908769635)
d’où
![{\displaystyle \operatorname {Cos} .P={\sqrt {-2{\sqrt {-1}}\operatorname {Sin} .x\operatorname {Cos} .x}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/37a2608755b4def150082a02c30cd989f317369f)
![{\displaystyle \operatorname {Cos} .Q={\sqrt {+2{\sqrt {-1}}\operatorname {Sin} .x\operatorname {Cos} .x}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5066061094d900910efede973cdebb0ca8b365c3)
donc
![{\displaystyle \operatorname {Cos} .P\operatorname {Cos} .Q=2\operatorname {Sin} .x\operatorname {Cos} .x\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0751b4322366d68386b161344875761f59e420d6)
on a d’ailleurs
![{\displaystyle \operatorname {Sin} .P\operatorname {Sin} .Q=1\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d3ac46a5b6a2ffd7982f2ac2a8812085d564c35f)
donc
![{\displaystyle \operatorname {Cos} .(P+Q)=\operatorname {Cos} .P\operatorname {Cos} .Q-\operatorname {Sin} .P\operatorname {Sin} .Q=2\operatorname {Sin} .x\operatorname {Cos} .x-1\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ccec2a8cc0d7d1cfc1d64cda0468abbc84c3961c)
donc aussi
![{\displaystyle P+Q=\operatorname {Arc} .(\operatorname {Cos} .=2\operatorname {Sin} .x\operatorname {Cos} .x-1),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c540819d0ff0bef44225d4369dd8b201c0b62cc)
et par suite
![{\displaystyle {\frac {P+Q}{2}}={\frac {1}{2}}\operatorname {Arc} .(\operatorname {Cos} .=2\operatorname {Sin} .x\operatorname {Cos} .x-1)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/27a5d17d76a2965ae15f5feb93b17fd81396ce21)
qui est conséquemment la somme finie demandée de la seconde des trois séries infinies proposées.
Si l’on suppose
on a
![{\displaystyle {\frac {P+Q}{2}}={\frac {1}{2}}\operatorname {Arc} .(\operatorname {Cos} .=-1)={\frac {1}{2}}\varpi \,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d0a8c27cd2ff2cf2e6fcb430c143966da1dcd2c)
donc, comme on le savait déjà,
![{\displaystyle {\frac {1}{2}}\varpi =1+{\frac {1}{2}}.{\frac {1}{3}}+{\frac {1.3}{2.4}}.{\frac {1}{5}}+{\frac {1.3.5}{2.4.6}}.{\frac {1}{7}}+{\frac {1.3.5.7}{2.4.6.8}}.{\frac {1}{9}}+\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/7703a4a2840c44cf8727d48b33f072523a0dd953)
Quant à la troisième série, on peut la mettre sous cette forme
![{\displaystyle {\frac {1}{2}}\left\{{\begin{array}{l}&{\frac {\operatorname {Cos} .(x+y)}{1}}-{\frac {\operatorname {Cos} .2(x+y)}{2}}+{\frac {\operatorname {Cos} .3(x+y)}{3}}-{\frac {\operatorname {Cos} .4(x+y)}{4}}+\ldots \\\\+&{\frac {\operatorname {Cos} .(x-y)}{1}}-{\frac {\operatorname {Cos} .2(x-y)}{2}}+{\frac {\operatorname {Cos} .3(x-y)}{3}}-{\frac {\operatorname {Cos} .4(x-y)}{4}}+\ldots \end{array}}\right\}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/84a4a95cc47fe3df740e59cd7a0007c0a21efd88)
Or on a, (pag. 114 du présent volume)
![{\displaystyle {\frac {\operatorname {Cos} .x}{1}}-{\frac {\operatorname {Cos} .2x}{2}}+{\frac {\operatorname {Cos} .3x}{3}}-{\frac {\operatorname {Cos} .4x}{4}}+\ldots =\operatorname {Log} .2+\operatorname {Log} .\operatorname {Cos} .{\frac {1}{2}}x,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bd22ec1c7ef0e5315ca129e2a455dbdd29ad7d58)
c’est-à-dire
![{\displaystyle {\frac {\operatorname {Cos} .x}{1}}-{\frac {\operatorname {Cos} .2x}{2}}+{\frac {\operatorname {Cos} .3x}{3}}-{\frac {\operatorname {Cos} .4x}{4}}+\ldots =\operatorname {Log} .2\operatorname {Cos} .{\frac {1}{2}}x\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/30ad345b64eb7fe8ec5c458fe0d5c18390717cf0)
en changeant donc, tour à tour
en
et
il viendra
![{\displaystyle {\frac {\operatorname {Cos} .(x+y)}{1}}-{\frac {\operatorname {Cos} .2(x+y)}{2}}+{\frac {\operatorname {Cos} .3(x+y)}{3}}-{\frac {\operatorname {Cos} .4(x+y)}{4}}+\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/b69e13bbcf07a0b4e2ae240d1a01e2905d9b851c)
![{\displaystyle =\operatorname {Log} .2\operatorname {Cos} .{\frac {1}{2}}(x+y)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7e0f5b3020351e7b18cd790b9699039161c6d4a1)
![{\displaystyle {\frac {\operatorname {Cos} .(x-y)}{1}}-{\frac {\operatorname {Cos} .2(x-y)}{2}}+{\frac {\operatorname {Cos} .3(x-y)}{3}}-{\frac {\operatorname {Cos} .4(x-y)}{4}}+\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/f99f6296f41a98aae4d650da754b82fd18016ff7)
![{\displaystyle =\operatorname {Log} .2\operatorname {Cos} .{\frac {1}{2}}(x-y)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/db127006ae26dc66c66c8219aca38555292bd20c)
donc, en prenant la demi-somme,
![{\displaystyle {\frac {\operatorname {Cos} .x\operatorname {Cos} .y}{1}}-{\frac {\operatorname {Cos} .2x\operatorname {Cos} .2y}{2}}+{\frac {\operatorname {Cos} .3x\operatorname {Cos} .3y}{3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4520745a7ac65ea9fa224dbcf7511dc3e90e8cbb)
![{\displaystyle -\ldots ={\frac {1}{2}}\left\{\operatorname {Log} .2\operatorname {Cos} .{\frac {1}{2}}(x-y)+\operatorname {Log} .2\operatorname {Cos} .{\frac {1}{2}}(x+y)\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ed0cb46eb8f1303338bddca144840120563425e5)
ou encore
![{\displaystyle {\frac {\operatorname {Cos} .x\operatorname {Cos} .y}{1}}-{\frac {\operatorname {Cos} .2x\operatorname {Cos} .2y}{2}}+{\frac {\operatorname {Cos} .3x\operatorname {Cos} .3y}{3}}-\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/cfb9de4f94011bc8d54ccb8a66cb49052719d0c0)
![{\displaystyle ={\frac {1}{2}}\operatorname {Log} .4\operatorname {Cos} .{\frac {1}{2}}(x-y)\operatorname {Cos} .{\frac {1}{2}}(x+y)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/655f5320db507282f632c220c4241d2ffda35b1e)
mais
![{\displaystyle 2\operatorname {Cos} .{\frac {1}{2}}(x-y)\operatorname {Cos} .{\frac {1}{2}}(x+y)=\operatorname {Cos} .x+\operatorname {Cos} .y}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a6248ea035124f97f6ec2a13c5c124f5c41ab96b)
d’où
![{\displaystyle 4\operatorname {Cos} .{\frac {1}{2}}(x-y)\operatorname {Cos} .{\frac {1}{2}}(x+y)=2\left(\operatorname {Cos} .x+\operatorname {Cos} .y\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f299c3c3ede56beebaf7202723304970a333e294)
donc enfin, la somme finie de la troisième des suites infinies proposées est
![{\displaystyle {\frac {1}{2}}\operatorname {Log} .\left(2\operatorname {Cos} .x+2\operatorname {Cos} .y\right).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fdd8a7fa31af9fd94b414b60ea01b1f3aa80c94c)
Si, au lieu de prendre la demi-somme des deux séries ci-dessus, on prend leur demi-différence, on aura
![{\displaystyle {\frac {\operatorname {Sin} .x\operatorname {Sin} .y}{1}}-{\frac {\operatorname {Sin} .2x\operatorname {Sin} .2y}{2}}+{\frac {\operatorname {Sin} .3x\operatorname {Sin} .3y}{3}}-\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/3a6c019aee42dc63565445aef58004da733f72e8)
![{\displaystyle ={\frac {1}{2}}\left\{\operatorname {Log} .2\operatorname {Cos} .{\frac {1}{2}}(x-y)-\operatorname {Log} .2\operatorname {Cos} .{\frac {1}{2}}(x+y)\right\}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/10dba92330916a201da61efbcb92fd9fdcba55c7)
ou encore
![{\displaystyle {\frac {\operatorname {Sin} .x\operatorname {Sin} .y}{1}}-{\frac {\operatorname {Sin} .2x\operatorname {Sin} .2y}{2}}+{\frac {\operatorname {Sin} .3x\operatorname {Sin} .3y}{3}}-\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/3a6c019aee42dc63565445aef58004da733f72e8)
![{\displaystyle ={\frac {1}{2}}\operatorname {Log} .{\frac {\operatorname {Cos} .{\frac {1}{2}}(x-y)}{\operatorname {Cos} .{\frac {1}{2}}(x+y)}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c3bb76b9e304a5bc5661cccdbcf784ef07c17f09)
Or,
![{\displaystyle {\frac {\operatorname {Cos} .{\frac {1}{2}}(x-y)}{\operatorname {Cos} .{\frac {1}{2}}(x+y)}}={\frac {2\operatorname {Cos} .{\frac {1}{2}}(x-y)\operatorname {Cos} .{\frac {1}{2}}(x+y)}{2\operatorname {Cos} .^{2}{\frac {1}{2}}(x+y)}}={\frac {\operatorname {Cos} .x+\operatorname {Cos} .y}{1+\operatorname {Cos} .(x+y)}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0f6e6a8267243e0fdb698f3548454519331a4834)
donc enfin
![{\displaystyle {\frac {\operatorname {Sin} .x\operatorname {Sin} .y}{1}}-{\frac {\operatorname {Sin} .2x\operatorname {Sin} .2y}{2}}+{\frac {\operatorname {Sin} .3x\operatorname {Sin} .3y}{3}}-\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/3a6c019aee42dc63565445aef58004da733f72e8)
![{\displaystyle ={\frac {1}{2}}\operatorname {Log} .{\frac {\operatorname {Cos} .x+\operatorname {Cos} .y}{1+\operatorname {Cos} .(x+y)}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e5a398201807860f873dbbc919d5fb114e4a25a7)
Si, dans ce résultat et dans le précédent, on fait
ils deviendront
![{\displaystyle {\frac {\operatorname {Cos} .^{2}x}{1}}-{\frac {\operatorname {Cos} .^{2}2x}{2}}+{\frac {\operatorname {Cos} .^{2}3x}{3}}-{\frac {\operatorname {Cos} .^{2}4x}{4}}+\ldots ={\frac {1}{2}}\operatorname {Log} .4\operatorname {Cos} .x,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6d3b405e28679eeb6a3a9fe7f2fa7beb209f6e69)
![{\displaystyle {\frac {\operatorname {Sin} .^{2}x}{1}}-{\frac {\operatorname {Sin} .^{2}2x}{2}}+{\frac {\operatorname {Sin} .^{2}3x}{3}}-{\frac {\operatorname {Sin} .^{2}4x}{4}}+\ldots ={\frac {1}{2}}\operatorname {Log} .{\frac {2\operatorname {Cos} .x}{1+\operatorname {Cos} .2x}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aa521fbbe342d8749cff3d6eb8a3f36de94a2b0b)
En résumé, si nous faisons abstraction des divers résultats particuliers auxquels nous sommes parvenus, et qui n’avalent pas été demandés, nous aurons
1.
o ![{\displaystyle {\frac {1}{2}}\operatorname {Arc} .\left(\operatorname {Tang} .={\frac {2a\operatorname {Cos} .x}{1-a^{2}}}\right)={\frac {a\operatorname {Cos} .x}{1}}-{\frac {a^{3}\operatorname {Cos} .3x}{3}}+{\frac {a^{5}\operatorname {Cos} .5x}{5}}-\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/90448bbf33aab5b94bd8388e4f45ebe31e236dbf)
2.
o ![{\displaystyle {\frac {1}{2}}\operatorname {Arc} .\left(\operatorname {Sin} .=2\operatorname {Sin} .x\operatorname {Cos} .x-1\right)=\operatorname {Cos} .x+{\frac {1}{2}}{\frac {\operatorname {Cos} .3x}{3}}+{\frac {1.3}{2.4}}.{\frac {\operatorname {Cos} .5x}{5}}+\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/4a329a791f00f8448436e0ca674e3d2fd7faf915)
3.
o ![{\displaystyle {\frac {1}{2}}\operatorname {Log} .2\left(\operatorname {Cos} .x+\operatorname {Cos} .y\right)={\frac {\operatorname {Cos} .x\operatorname {Cos} .y}{1}}-{\frac {\operatorname {Cos} .2x\operatorname {Cos} .2y}{2}}+{\frac {\operatorname {Cos} .3x\operatorname {Cos} .3y}{3}}-\ldots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/a4217e6d2c93c279b721a4b204897b6fadb899ef)
résultats qu’au surplus on peut présentement vérifier d’un grand nombre de manières diverses.