Plane trigonometry, by S.L. Loney.

COS 0 IN FACTORS. 441
In (4) make p infinite; then, as in the last article, we
have
cos0= l-44J[ 4l] 4 [2 -...2
[ 7r2 [12 - 32r2j 1 527r2j.. ad inf.
This theorem may be written in the form
r=oo 2
cos= II 1 40
r=l (2r -1)2 7r2
sin 20
Since cos 0 = sin 2' the product of cos 0 may be
derived from the products for sin 20 and sin 0.
371. The equation (4) of Art. 369 may, by means of Art. 362, be
shewn to be true for all integral values of p. For we have
2P - 2xp cospb + 1
= {2 - 2x cos 5+ 1} x2 - 2x cos (p + 2) + I)
2 - 2x cos ( + + 1......to p factors.
Put =1, and we have
2(1-cospq0)=-{2-2cos0} 2-22cos( ) t(0o f+-ac )...... to p factors.
i.e. 4 sin2 = 4 sin2 4 sin2( + ).4 sin2 ( + )...top factors.
Put 0=, and extract the square root of both sides. We have then
* 7.+0. 27r+6. (P-1)7r+0
+s in 0 = 2- sin. sin... sin0.....sin...(1).
P p p p
If 0 lie between 0 and 7r all the factors on the right-hand side of (1)
are positive and so also is sin 0. Hence the ambiguity should be
replaced by the positive sign.
If 0 lie between 7r and 27r, all the factors on the right-hand side are
positive except the last, which is negative.
Hence the product is negative and so also is sin 0, so that in this case
also the positive sign is to be taken.
Similarly in any other case it may be shewn that the positive sign
must be taken, and we have, for all integral values of p,
0. r+0. 27r+0. (p-1)}7+0
sin 0 = 2P- sin. si sin 0 +0 sin (
P. P P P