SECTION 1.3 Circle Geometry 39
proving the first part of Ptolemy’s theorem.
Assume, conversely, thatABCDis not cyclic, in which case it follows
that
CBÊ +ABĈ = CDÂ +ABĈ 6 = 180◦.
This implies that the pointsA, B,andE form a triangle from which
it follows thatEA < AB+BE.As above we apply (1.2) and (1.3) and
get
CA·DB
CD< AB+
CB·DA
CD
,
and so
CA·DB < AB·CD+CB·DA,
proving the converse.
Corollary. (The Addition Formulas for Sine and Cosine)We
have, for anglesαandβ, that
sin(α+β) = sinαcosβ+sinβcosα; cos(α+β) = cosαcosβ−sinαsinβ.
Proof. We shall draw a cyclic quadri-
lateral inside a circle having diameter
AC= 1 (as indicated), and leave the de-
tails to the reader. (Note that by Exer-
cise 3 on page 30, we have that BD =
sin(α+β) (see the figure). To obtain
the addition formula for cos, note that
cosα= sin(α+π/2).)