Plane trigonometry, by S.L. Loney.

106

TRIGONOMETRY.

106. An independent geometrical proof of the formulae
of the preceding article may be given, for values of A
which are less than a right angle.
Let QCP be the angle 2A.
With centre C and radius CP
describe a circle and let QC meet \ Q
it again in 0. \ N
Join OP and PQ, and draw PN
perpendicular to OQ.
By Euc. TI. 20, the angle
QOP = zQUCP=A,
and the angle NfPQ = QOP = A.
Hence
NP 2NP NP NP OP
sin 2A = 0 = 2 2 ' OQ
= 2 sin NOP cos POQ, since OPQ is a right angle,
= 2 sin A cos A;
also
CN 2CN (OC+ CN)-(OC-CN)
cos2A- =0-= 0 -ON-NQ ONOP NQPQ
OQ OP OQ PQ OQ
= cos2 A - sin2 A;
2NP
NP 2NP ON
and tan2A = =
O-N - oN- R P
PN 0N
2 tan A
1 - tan2 A'