FORMULES RELATIVES AUX FONCTIONS HYPERBOLIQUES
she Pubs 5
+ Chu — — Ë Thu= . ms À Ke [ v Séchu — Chu? Cothu = Tia RE À (Chu + Shu)" = Chru + Shru. 5
cos (iu) = Chu, Ch (iu) = cosu, Cho =#,
Chæ —, Ch(— u) = Chu,
aChu = Shu du,
d.Sh'u = d,.Ch?u = 2Shu Chu.du = Shou.du.
d.logShu = Cothu du,
dog Chu = Thu du,
tang (iu) = iThu. A
Th{iu) = étanigu. ne.
Tho — 0, 304
Theo =
Th(— Re Né
d "2 Thu = UE. Der
LE
dog Thu = ee
Shu= + Strass tt u° Che = 14 # LE , 12 1.3zx 13.547 AreShe= [ TES = log (x + x F)=È—i 215 ET RUE 2. — = log (x +7).
o Vz*=1 ArgThr= | = Log ET = + T+E+E nu ETS EE 5 o LE SP Me É 1 ArgCoth = Arg Th! a ne ue Rte to
ArgShz=ArgCh zx +1 = ArgTh—— = Fe
ArgChz = ArgShyxr?— 1 —
Pour x très-petit, |
2 de u = = cos Amhu — Shu = tangy,
Shu — u.(Chu}5,
Er
Thu = sing.
Run à, ce ide ct AS de EN a Et Dre SES
Thu =u.(Chu) 3, | = Shu.(Chu) 5 = Thu.(Chu}$, n:
®. x log tang (242).
9 = 2arc tange"— = Chu = sécy,
ArgTh! = —2ArgCh VE= 2ArgSh k
Ti u Cothu = 1.
Chu Séchu = 1.
Shu Coséchu — 1.
Séch?x = 1— Th'u.
Cosé hu — Coth?u — 1.
Thu
D Che
4 a PF Le >.
F1 # ne ’ | EU ne Pere « L - ’ 1. .
INTRODUCTION
Sh(u + e) = Shu Che + Chu She.
Ch{u + vw) = Chu Chr + Shu She.
Shu Che = £Sh(u+v)+1+Sh{(u —v).
Chu She = iSh(u+v)—+Sh(u—v).
Chu Cho — £Ch(u+ve)+£Ch(r —v).
Sh u Sho = £Ch(u+v)—+Ch(r—v).
Shu + She = 2Sh+{u+v)Ch+(r—v).
Sh & — She = 2Ch+(u+v)Shr(u — v).
Chu + Che = 2Ch4 (u + v)Ch+(« ef, Chu — Cho = 2Sh+{(u+v)Sh£(u —o).
Shu+ She Th, (u+v) Shu— She Thi(u—v) Chu + Che I Che Che Ths(u+e) Th#(u— 0) Shu + She Chu — Che Chu + Che Shw= She Thu +The
12 The The Cothw Cothe +1 À
= Thi(u +),
Th(u+vw) —
Shaw — 2Shu Chu. GRUEELTT Goihe & Cote Ch au = 2 Chu —1 Thu + Th — AE À ee = Ch°x + Sh’a Chu Che mn O —1+2Sh4. po Sh(ux#e) “4 De: Cothu + Cothe — Shr She Thau= ne, 1+Th'4 Ch(u+o) “0 ir d Cothu The = __ Cothe + Thu Sh? 4 — Sh?o = Ch'u — Ch°v = Sh (u+v)Sh (x — v). Cothau = — Ch?u+Sh°e = Sh?u+4+Ch?o = Ch(u+v)Ch(u —v). ae - 2Cotn “ ] ee : = 4Cothu + {Thu, Chu +1= 2Ch°tx, Cha — 1 = 2Sh°{w. hu 1+Thu Ms rte Shu __Chu—1 /Chu—:
Chu—Shz 1—The 7 Cothiu Chu+r Shu V Chu+1 - Cothu+i 1+4+Th'iu LAVTITT ET nd CES PEUTETE
Sh3u = 48h + 3Shu = Shu (4Ch?u — 1),
Sh (2 +1) u = 2Chu.Sh au — Sh(n —1)u, i C (a +s}u = 2Chu.Chru — Ch(n —1)u.
Des formules de la page xxxvi des 7”. à 5 D. il est aisé d’en déduire d’autres tives aux fonctions rboliques, en remplaçant sina, sin ra, CO8a, cos ra r iSha, iShra, Cha, Chra.
