Trignometric Series

Geometry Level 4

S = cos ( 2 π 28 ) csc ( 3 π 28 ) + cos ( 6 π 28 ) csc ( 9 π 28 ) + cos ( 18 π 28 ) csc ( 27 π 28 ) S = \cos \left( \frac{2 \pi}{28} \right) \csc \left( \frac{3 \pi}{28} \right) + \cos \left( \frac{6 \pi}{28} \right) \csc \left( \frac{9 \pi}{28} \right) + \cos \left( \frac{18 \pi}{28} \right) \csc \left( \frac{27 \pi}{28} \right)

Find the value of log 2 3 ( 1 + S + S 2 ) \left| \log_{2 - \sqrt{3}} \left( 1 + S + S^{2}\right)\right| .


The answer is 0.

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2 solutions

Avi Solanki
Feb 10, 2016

Let A=pi/28 given expression can be written as

cos2A/sin3A + cos6A/sin9A + cos18A/sin27A

  • cos2A=sin12A

cos18A= -sin4A

cos6A=sin8A

sin27A=sinA*

*1) cos2A/sin3A= sin12A/sin2A =(4sin3A.cos3A.cosA) /sin3A = 4cos3Acos6A

2)cos18A/sin27A= (-sin4A/sinA)= - (4sinAcosA.cos2A)/sinA= -(4cosAcos2A)

(1)+(2)= 2cos9A-2cosA

Now [2cos9A -2cosA] +cos6A/sin9A = (sin18A-2cosA.sin9A +cos6A)/sin9A *

= *(sin18A-sin8A +sin8A -sin10A)/sin9A =(2cos14A.sin4A)/sin9A [cos14A=cosPi/2=0] =0 *
S=0. therefore the expression is log 1=0 .

Nice thinking ! I have one more way

Jaideep Khare - 5 years, 4 months ago

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Nice solution .!!

avi solanki - 5 years, 4 months ago
Zerocool 141
Feb 10, 2016

very nice solution .the conversion was the main trick to this problem. upvoted . :)

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