Question #cb7c5

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Question #cb7c5

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Answer 1 · 2016-07-04T19:07:12

$= \frac{1}{32}$ $=1/32$

Explanation:

$cos (\frac{π}{11}) cos (2 \frac{π}{11}) cos (3 \frac{π}{11}) cos (4 \frac{π}{11}) cos (5 \frac{π}{11})$ $cos(pi/11)cos(2pi/11)cos(3pi/11)cos(4pi/11) cos(5pi/11)$

$= \frac{4 sin (\frac{π}{11}) cos (\frac{π}{11}) cos (2 \frac{π}{11}) cos (3 \frac{π}{11}) cos (4 \frac{π}{11}) cos (5 \frac{π}{11})}{4 sin (\frac{π}{11})}$ $=(4sin(pi/11)cos(pi/11)cos(2pi/11)cos(3pi/11)cos(4pi/11)cos(5pi/11))/(4sin(pi/11))$

$= \frac{2 sin (4 \frac{π}{11}) cos (4 \frac{π}{11}) cos (π - \frac{8pi}{11}) cos (5 \frac{π}{11})}{8 sin (\frac{π}{11})}$ $=(2sin(4pi/11)cos(4pi/11)cos(pi-"8pi"/11)cos(5pi/11))/(8sin(pi/11))$

$= - \frac{2 sin (8 \frac{π}{11}) cos (8 \frac{π}{11}) cos (5 \frac{π}{11})}{16 sin (\frac{π}{11})}$ $=-(2sin(8pi/11)cos(8pi/11)cos(5pi/11))/(16sin(pi/11))$

$= - \frac{2 sin (16 \frac{π}{11}) cos (5 \frac{π}{11})}{32 sin (\frac{π}{11})}$ $=-(2sin(16pi/11)cos(5pi/11))/(32sin(pi/11))$

$= - \frac{2 sin (π + \frac{5pi}{11}) cos (5 \frac{π}{11})}{32 sin (π - \frac{π}{11})}$ $=-(2sin(pi+"5pi"/11)cos(5pi/11))/(32sin(pi-pi/11))$

$= \frac{sin (10 \frac{π}{11})}{32 sin (10 \frac{π}{11})}$ $=(sin(10pi/11))/(32sin(10pi/11))$

$= \frac{1}{32}$ $=1/32$

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

$Multiplyig both numerator and denominator by 4 sin (\frac{π}{11})$ $"Multiplyig both numerator and denominator by " 4sin(pi/11)$

$Applying identity 2 sin A cos A = sin 2 A$ $"Applying identity "2sinAcosA=sin2A$

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Question #cb7c5

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