Of KEPLER'S PROBLEM. 219- 



the fame co-efficient, fubftitute the produdls that are equal to 



them ; we fhall have, 



e fin 2<p cof lip =z 2 fin « fin <p fin 4/ + 2 cof « cof (p fin -^^ and 



e cof 2(p fin 2-4/ ::i — 2 fin 7i coi'ip cof i|/ -}- 2 cof« fin (p cof •»//. 

 Now, fin « fin (p + cof n cof (p =: cof (<p — «), and 



— fin n cof <p + cof « fin (p :::: fin ((p — «) : therefore, 

 e fin 2ip cof 2-4/ r: 2 cof ((p — n) fin ■^~\ . . x 

 (? cof 2(p fin 2-^ — 2 fin ((p — 77) cof -^/J ^ ' 



If in the fecond of the equations (A), we write 2 fini|/ cof^/ for 

 fin 2>J/, and divide by cofi|/, we fiiall obtain f cof 2(p fin t|/ — fin ((p— «); 



whence fin 4' = - X ^-^^^ —. But cof 2 J/ =: i — 2 fin '-vL =: 



f coi i(p ^ ^ 



I X 7^ : Subftitute thefe values of fin 4 and 



e'^ col ' ii(p 



cof 24' in the firft of the equations (A), and, after having redu- 

 ced, there will refult, 

 £^fin2(pcof^2(p — 2fin'((p — 7;)fin2<pz: 2fin((p — 7/)cof((p — «) cof 2(p.. 

 Now, 2 fin " (<p — ») '=■ I — cof (2(p — 2«)i and 2 fin (<p — «) 

 X cof ((p — ?-!) = fin (2(p — 2n) : therefore, by properly ordering 

 the terms, 



fin2(p— (?^fin2(pcof*2<p=:fin 2<pcof(2(p — 2n)— cof 2ipfin(2(p— 2ff). 

 But fin 2(p cof {2<p — 2») -7- cof 2ip fin (2<p — 2h) :=. fin 2« : 

 therefore, if we put x — fin 2<p, fince cof* 2<p zz i — a?*, our 

 equation will finally become 



- , /i ^\ fin 2« 



which equation will ferve to determine x rz fin 2<p. 



We have ftiU to find the angle 4/ : for this purpofe I refume 

 the equations (A), and multiplying crofs-wife, and dividing by 

 2e, there refults, 



fin 2(p fin {<p — n) cof 24 cof 4- r: cof 2(p cof (<p — n) fin 24 fin 4. 

 For fin 24 write 2 fin 4- cof <]> : and for 2 fin "^ 4 write i — cof 24 ; 

 then, having properly difpofed the terms, we fliall find 

 (iin 2<p fin (^(p — n)-\- cof 2^ cof (cp— »)) cof 24 =: cof 2(p cof ((p— n). 



Now 



