280 
TRIGONOMETRY 
The equation in fact gives the formula, 
{crd. (a — /3)}. (crd. 180°) = (crd. a). {crd. (180° —/3)} 
— (crd. (3). {crd. (180° — ex)}, 
which is, of course, equivalent to 
sin (0 — 0) = sin 0 cos 0 —cos 0 sin 0, where a = 2 0, ¡3 = 20. 
By means of this formula Ptolemy obtained 
crd. 12° = crd. (72°-60°) = 12P 32' 36". 
(8) Equivalent of sin 2 i0 = \ (1 —cos0). 
But, in order to get the chords of smaller angles still, we 
want a formula for finding the chord of half an arc when the 
chord of the arc is given. This is the subject of Ptolemy’s 
next proposition. 
Let BG be an arc of a circle with diameter AC, and let the 
arc BG be bisected at D. Given (crd. BG), it is required to 
find (crd. BG). 
Draw DF perpendicular to AC, 
and join AB, AD, BD, DC. Measure 
AE along AG equal to AB, and join 
BE. 
Then shall FG be equal to EF, or 
FG shall be half the difference be 
tween AG and AB. 
For the triangles ABB, AEB are 
equal in all respects, since two sides 
of the one are equal to two sides of the other and the included 
angles BAD, EAD, standing on equal arcs, are equal. 
Therefore ED — BB = BG, 
and the right-angled triangles BEF, BGF are equal in all 
respects, whence EF = FG, or CF — ±{AG—AB). 
Now AG. GF = CD 2 , 
whence (crd. CD) 2 = \AG {AG— AB) 
= i(crd. 180°). {(crd. 180°) - (crd. 180°-50)}. 
This is, of course, equivalent to the formula