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prove that cos(pi/15).cos(2pi/15).cos(3pi/15).cos(4pi/15).cos(5pi/15).cos(6pi/15).cos(7pi/15)=1/128

sujith , 7 Years ago
Grade 11
anser 2 Answers
Vikas TU

Last Activity: 7 Years ago

LHS = cos(pi/15).cos(2pi/15).cos(3pi/15).cos(4pi/15).cos(5pi/15).cos(6pi/15).cos(7pi/15) 
Duplicate and partition LHS by 128sin(pi/15) 
We realize that 2 sin a cos a = sin 2a 
=> {64/128 sin (pi/15) } . 2 sin (pi/15) . cos(pi/15) .cos(2pi/15) .cos(3pi/15) .cos(4pi/15) .cos(5pi/15).cos(6pi/15).cos(7pi/15) 
=> {64/128 sin (pi/15) }.sin(2pi/15) .cos(2pi/15) .cos(3pi/15) .cos(4pi/15) .cos(5pi/15) .cos(6pi/15).cos(7pi/15) 
=> {32/128 sin (pi/15) }.sin(4pi/15) .cos(3pi/15) .cos(4pi/15) .cos(5pi/15) .cos(6pi/15).cos(7pi/15) 
=> {16/128 sin (pi/15) }sin (8pi/15) .cos(3pi/15) .cos(5pi/15).cos(6pi/15).cos(7pi/15) 
=> {16/128 sin (pi/15) } sin (7pi/15) .cos(3pi/15) .cos(5pi/15).cos(6pi/15).cos(7pi/15) 
[since sin(8pi/15) can be composed as sin(pi – 8pi/15) = sin(7pi/15) ] 
Also making each of the 2 sin a cos a = sin 2a 
=1/128. 

Bizhan assadian

Last Activity: 7 Years ago

First method (use Formula):
There is a simple, yet obscure trig formula, which solves this problem instantly. Here it is:
cos(pi/2k+1) cos(2pi/2k+1) cos(3pi/2k+1).....cos(kpi/2k+1) = 1/2^k
For this problem let 2k + 1 = 15 ---> k = 7 , thus the answer = 1/2^7 = 1/128
Bizhan Assadian

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