S = cos ( 2 8 2 π ) csc ( 2 8 3 π ) + cos ( 2 8 6 π ) csc ( 2 8 9 π ) + cos ( 2 8 1 8 π ) csc ( 2 8 2 7 π )
Find the value of ∣ ∣ lo g 2 − 3 ( 1 + S + S 2 ) ∣ ∣ .
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Nice thinking ! I have one more way
very nice solution .the conversion was the main trick to this problem. upvoted . :)
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Let A=pi/28 given expression can be written as
cos2A/sin3A + cos6A/sin9A + cos18A/sin27A
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 .