PTOLEMY’S 8YNTAXI8 
281 
By successively applying this formula, Ptolemy obtained 
(crd. 6°), (crd. 3°) and finally (crd. lf°) = IP 34' 15" and 
(crd. |°) = OP 47' 8". But we want a table going by half 
degrees, and hence two more things are necessary; we have to 
get a value for (crd. 1°) lying between (crd. lf°) and (crd. |°), 
and we have to obtain an addition formula enabling us when 
(crd. a) is given to find (crd. (a + ^°)}, and so on. 
(e) Equivalent of cos (0 + 0) = cos 0 cos 0 —sin 0 sin 0. 
To find the addition formula. Suppose AD is the diameter 
of a circle, and AB, BG two arcs. Given (crd. AB) and 
(crd. BG), to find (crd. AG). Draw the diameter BOE, and 
join GE, GD, DE, BD. 
Now, (crd. AB) being known, 
(crd. BD) is known, and therefore 
also (crd. DE), which is equal to 
(crd. AB)-, and, (crd. BG) being- 
known, (crd. GE) is known. 
And, by Ptolemy’s theorem, 
BD. GE = BG. DE + BE. GD. 
The diameter BE and all the chords in this equation except 
GD being given, we can find CD or crd. (180° — AG). We have 
in fact 
(crd. 180°). {crd. (180°- AG)] 
—[crd. (180° — AB)}. {crd. (180° — BG)] — {crd.AB). (crd. BG); 
thus crd. (180° — AG) and therefore (crd. AG) is known. 
If AB = 2 6, BG =2 0, the result is equivalent to 
cos (0 + 0) = cos 0 cos 0 — sin 0 sin 0. 
(/) Method of interpolation based on formula 
sin a /sin < oi/(3 {where \ n > a > /3). 
Lastly we have to find (crd. 1°), having given (crd. 1^°) and 
(crd. |°). 
Ptolemy uses an ingenious method of interpolation based on 
a proposition already assumed as known by Aristarchus. 
If AB, BG be unequal chords in a circle, BG being the