A history of Greek mathematics - Wilbourhall.org

PTOLEMY'S tiYNTAXIS 281
By successively applying this formula, Ptolemy obtained
(crd. 6°), (crd. 3°) and finally (crd. 1|°) = 1* 34' 15" and
(crd. |°) = OP 47' 8". But we want a table going by halfdegrees,
and hence two more things are necessary ; we have to
get a value for (crd. 1°) lying between (crd. 1|°) and (crd. f °),
and we have to obtain an addition formula enabling us when
(crd. a) is given to find {crd. (a + J°)}, and so on.
i
(e) Equivalent of cos (0 + (p) = cos 6 cos — sin 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 CE, 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. CE) is known.
And, by Ptolemy's theorem,
BD . CE DE+BE. GD.
= BG .
The diameter BE and all
the chords in this equation except
We have
GD being given, we can find CD or crd. (180° — A C).
in fact
(crd. 180°) .
{crd. (180° -AC)}
= {crd. (180° - AB) \.{cvd. (180° -BC)} -(crd. AB). (crd. BC);
thus crd. (180° — AC) and therefore (crd. AC) is known.
If AB = 2 0, BG = 2 0, the result is equivalent to
cos (0 + (f>)
= cos cos — sin 6 sin 0.
(() Method of interpolation based on formula
sin oc /sin (3 < oi/fi
Lastly we have to find (crd.
(where \ tt > oc > ft).
1°), having given (crd. 1J°) 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