Of KEPLER'S PROBLEM. 2ri 



From the equation cof2v= — ^1 — ~ — y'^^ readily obtain 



, _ ^ (cof ^ (n — sr) — cof ^ 257-) ^fin ^ 27r — fin '' (/; — g- ) 



"° 2" — cof (« — tt) cof (« — ^) ^ 



but fin (« — t) r: ^ fin 2r, therefore, by fubftitution, 



r fin 2t ^^ /I '? 



cof(« — ff) '^ 



Multiply both fides by - ; and fubflitute, for ^ fin av, its equal 



« — 11 ; and for ~ fin a^^-, its equal fin (« — ir), and we fl^iall 

 have. 



fin (n — t") 



cof (/i — ^) ^ "^ ^ 4 ' 



Subtradl both fides of this equation from n — •r, and,, remark- 



1 r ' \ fin (// — ■^") ■ ... ^ , 



mg that tan [n — ■tt) — ^ ^ there will refult. 



"j: zz. tt — It — tan (« — -x) x J ^ — - ■ 



4 



To fimplify this formula, I put » — ^ = f > and a zz kJ i — ~: 



and fo, (in the cafe when v — •y is a maximum) we have 

 V — TT zz ^ — ^ tan f. 

 Let us now confider the funflion ^ — a tan ^ : becaufe 



n zz, sj I — - is lefs than unity, it is evident that we may take 



the arch ^ fo fmall, that a tan ^ fliall be lefs than ^ ; and, that 

 there is a value of g fuch that ^ zz a tan ^ : it is alfo manifeft, 

 that between the limits f = o and ^zz a tan f, (in both which 

 cafes ^ — a tan ^ zz 6) the funeftion ^ — n tan ^ attains a max- 

 imum value. 



If, therefore, we can prove, -that the arch n — t is always 

 between the limits f = o and f = ^ tan ^ ; it will follow that 

 the maximum value of the fundtion ^ — a tan ^ is greater than 

 tlie. arch v — %, even when that arch is greateft of all. 



Now,. 



