Of KEPLER'S PROBLEM. 219 



the fame coefficient, fubftitute the produ&s that are equal to 



them ; we fhall have, 



e fin 2<p cof 2i|/ = 2 fin n fin <p fin -^ + 2 cof « cof <p fin 4", and 



e cof2(p fin 2^ rr — 2 fin » cof <p cof -^ -f- 2 cof« fin <p cof -vj/. 

 Now, fin n fin <p -f- cof « cof <p =± cof (<p — «), and 



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

 e fin 2<p cof 2-^ — 2 cof (<p — ?i) fin ^\ r\\ 

 e cof 2<p fin 2-^ =: 2 fin (<p — ») cof i£ J 



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



fin 2i|/, and divide by cof -4/, we lhall obtain if cof 2<p fin ^ — fin (<p— »)j 



whence fin 4 := - X 1 -^~ — . But cof iA> — 1 — 2 fin ^ — 



e cof 2<p 



1 ? x — — y a ' : Subflitute thefe values of fin ^ and 



^ 2 cof 2 2<p 



cof 2^ in the firft of the equations (A), and, after having redu- 

 ced, there will refult, 

 ^fin2<pcof*2<p — 2fin 1 ((p~-«)fin2(p = 2fin((p— n)co{(<p— »)cof2<p< 



Now, 2 fin 2 (<p — ») = 1 — cof (2<p — 2»), and 2 fin (p — «) 

 X cof (<p — ») =. fin (2<p — 2«) :. therefore, by properly ordering 



the terms, 



fin2(p— <? 2 fin2<pcof 2 2<p:=:fin2<pcof(2<p — 2ri) — cof 2<pfin(2<£>— 2ff). 



But fin 2(p cof (2<p — 2») — cof 2<p fin (2<p — in) = fin 2» : 

 therefore, if we put * £= fin 2<p, fince cof 1 2<p = I — x\ our 

 equation will finally become 



/ 1 \ fin 2n 



x +K?- T ) X ^~^' 



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



We have ftill to find the angle -^ : 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 2^ cof 4- = cof 2<p cof (<p — n) fin 2^ fin 4*. 

 For fin 2^ write 2 fin ^ cof ^ : and for 2 fin 2 4/ write 1 — cof 2^ ; 

 then, having properly difpofed the terms, we fhall find 

 (iin 2$ fin ^<p — n)-\- cof 2<pcof (<p— «)) cof 2^ = cof 2<pcof(<p — »), 



Now 



