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

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Arun · Answer

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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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)=

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