A history of Greek mathematics Vol.II from Aristarchus to Diophantus by Heath, Thomas Little, Sir, 1921

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.
(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
i
282 TRIGONOMETRY
greater, then shall the ratio of CB to BA be less than the
ratio of the arc CB to the arc BA.
Let BD bisect the angle ABC, jneeting AC in E and
Now
Componendo, FA :
the circumference in D. The arcs
AD, DC are then equal, and so are
the chords AD, DC. Also CE>EA
(since CB:BA = CE:EA).
Draw DF perpendicular to AC
;
then AD>DE>DF, so that the
circle with centre D and radius DE
will meet DA in G and DF produced
in#.
FE: EA = AFED : AAED
< (sector HED) : (sector GED)
< IFDEiAEDA.
AE < Z FDA : Z ^D^.
Doubling the antecedents, we have
CA:AE < LCDA-.LADE,
and, separando, CE:EA < Z CDE: Z EDA
;
therefore
i. e. (crd. CB) :
(since CB:BA = CE:EA)
CB.BA < ICDBUBDA
< (arc CB): (arc BA),
(crd. 5,4) < (arc CB) :
(arc BA ).
[This is of course equivalent to sin oc : sin fi < a :
/3, where
i7i->a>/3.]
It follows (1) that (crd. 1°) : (crd. j°)< 1 :|,
and (2) that (crd. li°) : (crd. 1°) < If
: 1.
That is,
-f
. (crd. |°) > (crd. 1°) > |
. (crd. 1|°).
But (crd. |°) = OP 47' 8", so that f (crd. |°) = IP 2' 50"
nearly (actually IP 2' SOf")
and (crd. lj°) = 1* 34' 15", so that |(crd. lf°) = IP ^ 50".
Since, then, (crd. 1°) is both less and greater than a length
which only differs
inappreciably from IP 2' 50", we may say
that (crd. 1°) — \P 2' 50" as nearly as possible.
,