Geometry and Trigonometry 671
Using the identity sina−cosa=
√
2 sin(a− 45 ◦)in the numerators, we transform this
further into
√
2 sin( 1 ◦− 45 ◦)·
√
2 sin( 2 ◦− 45 ◦)···
√
2 sin( 44 ◦− 45 ◦)
sin 1◦sin 2◦···sin 44◦
=
(
√
2 )^44 (− 1 )^44 sin 44◦sin 43◦···sin 1◦
sin 44◦sin 43◦···sin 1◦
.
After cancellations, we obtain 2^22.
696.We can write
√
3 +tann◦=tan 60◦+tann◦=
sin 60◦
cos 60◦
+
sinn◦
cosn◦
=
sin( 60 ◦+n◦)
cos 60◦cosn◦
= 2 ·
sin( 60 ◦+n◦)
cosn◦
= 2 ·
cos( 30 ◦−n◦)
cosn◦
.
And the product telescopes as follows:
∏^29
n= 1
(
√
3 +tann◦)= 229
∏^29
n= 1
cos( 30 ◦−n◦)
cosn◦
= 229 ·
cos 29◦cos 28◦···cos 1◦
cos 1◦cos 2◦···cos 29◦
= 229.
(T. Andreescu)
697.(a) Note that
1 −2 cos 2x= 1 − 2 (2 cos^2 x− 1 )= 3 −4 cos^2 x=−
cos 3x
cosx
.
The product becomes
(
−
1
2
) 3
cos^37 π
cosπ 7
·
cos^97 π
cos^37 π
·
cos^277 π
cos^97 π
=−
1
8
·
cos^277 π
cosπ 7
.
Taking into account that cos^277 π=cos( 2 π−π 7 )=cosπ 7 , we obtain the desired identity.
(b) Analogously,
1 +2 cos 2x= 1 + 2 ( 1 −2 sin^2 x)= 3 −4 sin^2 x=
sin 3x
sinx
,
and the product becomes
1
24
sin^320 π
sin 20 π
·
sin^920 π
sin^330 π
·
sin^2720 π
sin^920 π
·
sin^8120 π
sin^2720 π
=
1
16
sin^8120 π
sin 20 π