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The value of cos(pi/15)cos(2pi/15)cos(3pi/15)cos(4pi/15)cos(5pi/15)cos(6pi/15)cos(7pi/15)=

The value of cos(pi/15)cos(2pi/15)cos(3pi/15)cos(4pi/15)cos(5pi/15)cos(6pi/15)cos(7pi/15)=

Grade:11

1 Answers

Arun
25758 Points
4 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. 

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