the excentric from the mean anomaly of a planet. 131 
Hence R° z = 
R° 
I ~f“ f/ 
e *4“ 
R°a 
2(i+ciy 
™ 4 .LW_,. + fc 
e '+6Wr e ^ 
2(l+rf) s 
Second Method. 
The substitutions for the sines and cosines of the angles 
CMS, SMN being as in the preceding method, let SG be per- 
pendicular to MC, and then radius being 1, 1 : CS : : sin. 
ACM : CS x sin, ACM= SG. But sin. CMS : SG : : sin. SMN : 
SN, that is, a : CS x sin. ACM : : s : 
CS x sin. ACM 
s = SN, and 
therefore ac — bs + 
CS x sin. ACM 
CS x sin. ACM 
.9 = TN 4- SN = ST 
r* b, and then ac + ds 
CMS — z. Let d • 
CMS — z 9 and CMS == % 4 " ac 4 ” ds 3 as i n the preceding 
method. 
Third Method. 
Let % = the angle SMN, 5 = the series expressing its sine, 
and c — the series expressing its cosine, as before; but let a 
now denote the sine of CSM, and b its cosine, and let MN 
meet CS in H. Then ac 4 - bs is equal to the sine of the sum 
of the angles SMH, MSC, that is, ac 4- bs = sin. MHA = sin. 
ACE, the excentric anomaly. We have therefore CE : ac 4- bs 
. . eg ; +±1 — ST = CMS — 
Let d = CS f a , and e = ~™,and then dc 4- £9— CMS — z. 
CE 
dz 2 
and CMS ~ z 4 “ dc 4 ” es — z d —■ — - — 4 ~ 
dz* 
dz , 6 
4- &c. 4 - e% 
4, ±1+ J±- 
" 2.3.4 1 2.34.5 
ez 3 
ez s 
4 “ 
2.3 1 2. 3.4.5 
2 -34 
. — &c. ^2= d 4~ z 4" ez * 
dz 2 
2 
2.34.5.6 
ez? 
2-3 
&c. Let/ = CMS — d, and then/ == Az 
— Bz 2 — Cz 3 4* Dz 4 4- Ez 5 — &c. putting A, B, C, &c. for the 
coefficients. By reversing this series, or by putting d for a, 
S 2 
