670 Geometry and Trigonometry
nlim→∞Rn=∏∞
n= 1cosπ
2 n.
The product can be made to telescope if we use the double-angle formula for sine written
as cosx=2 sinsin 2xx. We then have
∏∞
n= 2cosπ
2 n
= lim
N→∞∏N
n= 2cosπ
2 n
= lim
N→∞∏N
n= 21
2
·
sin
π
2 n−^1
sinπ
2 n= lim
N→∞1
2 N
sin
π
2
sinπ
2 N=
2
πlim
N→∞π
2 N
sinπ
2 N=
2
π.
Thus the answer to the problem isπ^2.
Remark.As a corollary, we obtain the formula
2
π=
√
2
2
·
√
2 +
√
2
2
·
√
2 +
√
2 +
√
2
2
···.
This formula is credited to F. Viète, although Archimedes already used this approximation
of the circle by regular polygons to computeπ.
694.Fork= 1 , 2 ,...,59,
1 −
cos( 60 ◦+k◦)
cosk◦=
cosk◦−cos( 60 ◦+k◦)
cosk◦=
2 sin 30◦sin( 30 ◦+k◦)
cosk◦
=
cos( 90 ◦− 30 ◦−k◦)
cosk◦=
cos( 60 ◦−k◦)
cosk◦.
So
∏^59k= 1(
1 −
cos( 60 ◦+k◦)
cosk◦)
=
cos 59◦cos 58◦···cos 1◦
cos 1◦cos 2◦···cos 59◦= 1.
695.We have
( 1 −cot 1◦)( 1 −cot 2◦)···( 1 −cot 44◦)=(
1 −
cos 1◦
sin 1◦)(
1 −
cos 2◦
sin 2◦)
···
(
1 −
cos 44◦
sin 44◦)
=
(sin 1◦−cos 1◦)(sin 2◦−cos 2◦)···(sin 44◦−cos 44◦)
sin 1◦sin 2◦···sin 44◦