Of KEPLER'S PROBLEM. 217 



Also, fince fin 2v — — ; tt-t- ; we have i — fin 2v zz 



' I + col A 



i-cofA ^"'1 A 



I + col-A = ^=tan»-: now let 2.-/. = 9o - 2X ; 



coP — 



2 



then I . — fin 2^ =: T — cof 2X =; 2 fin^ A : therefore fin X- rr 

 tan - X -y: = tan -- X fin 45°. 



2 v/2 2 



Inverting this analyfis, we derive the following rule for 

 computing the eccentric anomaly, when the mean anomaly 

 is a right angle: Take tan A := ^ X fec45° ; then fin X rx 



A • ' . 



tan — X fin 45°, and ^ =r 90° — 2X. This rule would be ri- 

 gorous and exadt, if we could give to e the value it has in the 



m — y. 

 fill — 7 — 

 form\ila f =: sX — ■ but as this cannot be done, we 



muft be content with approximating to the arch fought as near- 

 ly as our purpofe may require. We will, therefore, by means 

 of the rule, compute aferiesof arches, p,p',p'', &c. by fucceffively 



fin ——iL f^n ^ 



2 2 



fubftituting for e, the values s, i x -, s X , &c. 



^im—p)- ~[m—p') 



and the feries p, p\ p", &c. will converge very quickly to the. 

 exadl value of the arch of eccentric anomaly, erring alternately 

 in defeifl and in excefs. For the arches here denoted by p, p'., . 

 p", &-€. are manifeftly no other than the double of the arches 

 formerly denoted by v, ff', tr", &<:. refpedlively. 



The cafe of the general problem that we have here refolved, 

 may be thus enunciated : " To draw a ftraight line from aa 



" eccentric 



