Of KEPLER'S PROBLEM. 211 



From the equation cof 2v = — p- — v, we readily obtain 



1 col [n — n) J 



~ ^/ (cof % (a — <?r} — cof 2 2ct) y/fin 2 2x — fin * (a — sr ) 



lin 2 " — cof (a — ?r) cof O — a-) > 



but fin (« — t) n: - fin 2#, therefore, by fubftitution, 



r fin 2T . . /T"~ e* 



fin 2v= — ^ rXV 1 -;. 



Multiply both fides by ~ ; and fubflitute, for '- fin 2t>, its equal 



E 



2 



« — r; and for ~ fin 2t, its equal fin (n — ^), and we fhall 



have, 



fin (n — ?r) /t ^ 



cof (« — t) ^ 4 



Subtract both fides of this equation from n — t, and > remark- 



i / ' \ fin (n — v) ' - _ , 



mg that tan (« — v) ~ —^ { there will refult, 



9 cof [n — 5r)' ' 



— w z= » — # — tan (« — a-) x \f r — -• 



To fimplify this formula, I put n — v rr g , and a — sj 1 — L. : 



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



v — ■ or ~ g — a tan g. 

 Let us now confider the function § — a tan § : becaufe 



a — hj 1 — - is lefs than unity, it is evident that we may take 



the arch g fo fmall, that a tan g fhall be lefs than g ; and, that 

 there is a value of § fuch that g — a tan g : it is alfo manifefl, 

 that between the limits g =. o and f — # tan g>, (in both which 

 cafes ^ — # tan g> r: o) the function g — a tan g attains a max- 

 imum value. 



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

 between the limits f — o and g — a tan £ ; it will follow that 

 the maximum value of the function g — a tan § is greater than 

 tire arch v — t, even when that arch is greateft of all. 



Now, 



