Advanced book on Mathematics Olympiad

672 Geometry and Trigonometry

Because sin^8120 π=sin( 4 π+ 20 π)=sin 20 π, this is equal to 161.

(T. Andreescu)

698.(a) We observe that

secx=

1

cosx

=

2 sinx

2 sinxcosx

= 2

sinx

sin 2x

.

Applying this to the product in question yields

∏^24

n= 1

sec( 2 n)◦= 224

∏^24

n= 1

sin( 2 n)◦

sin( 2 n+^1 )◦

= 224

sin 2◦

sin( 225 )◦

.

We want to show that sin( 225 )◦=cos 2◦. To this end, we prove that 2^25 − 2 −90 is

an odd multiple of 180. This comes down to proving that 2^23 −23 is an odd multiple

of 45= 5 ×9. Modulo 5, this is 2·( 22 )^11 − 3 = 2 ·(− 1 )^11 − 3 =0, and modulo 9,

4 ·( 23 )^7 − 5 = 4 ·(− 1 )^7 − 5 =0. This completes the proof of the first identity.

(b) As usual, we start with a trigonometric computation

2 cosx−secx=

2 cos^2 x− 1

cosx

=

cos 2x

cosx

.

Using this, the product becomes

∏^25

n= 2

cos( 2 n+^1 )◦

cos( 2 n)◦

=

cos( 226 )◦

cos 4◦

.

The statement of the problem suggests that cos( 226 )◦=−cos 4◦, which is true only if

226 −4 is a multiple of 180, but not of 360. And indeed, 2^26 − 22 = 4 ( 224 − 1 ), which is

divisible on the one hand by 2^4 −1 and on the other by 2^6 −1. This number is therefore

an odd multiple of 4× 5 × 9 =180, and we are done.

(T. Andreescu)