Advanced book on Mathematics Olympiad

(ff) #1
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

.
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