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Answer 1 · 2016-07-04T19:07:12

$=1/32$

$cos(pi/11)cos(2pi/11)cos(3pi/11)cos(4pi/11) cos(5pi/11)$

$=(4sin(pi/11)cos(pi/11)cos(2pi/11)cos(3pi/11)cos(4pi/11)cos(5pi/11))/(4sin(pi/11))$

$=(2sin(4pi/11)cos(4pi/11)cos(pi-"8pi"/11)cos(5pi/11))/(8sin(pi/11))$

$=-(2sin(8pi/11)cos(8pi/11)cos(5pi/11))/(16sin(pi/11))$

$=-(2sin(16pi/11)cos(5pi/11))/(32sin(pi/11))$

$=-(2sin(pi+"5pi"/11)cos(5pi/11))/(32sin(pi-pi/11))$

$=(sin(10pi/11))/(32sin(10pi/11))$

$=1/32$

$color(red)("Explanation of steps")$

$"Multiplyig both numerator and denominator by " 4sin(pi/11)$

$"Applying identity "2sinAcosA=sin2A$