Ifcos⁡π7cos⁡2π7cos⁡4π7=sin⁡8π78sin⁡π7 then, sin⁡π14sin⁡3π14sin⁡5π14sin⁡7π14sin⁡9π14sin⁡11π14sin⁡13π14=

=sin⁡π14sin⁡3π14sin⁡5π142=−cos⁡π7cos⁡2π7cos⁡4π72=sin⁡8π78sin⁡π72=164

Ifcos⁡π7cos⁡2π7cos⁡4π7=sin⁡8π78sin⁡π7 then, sin⁡π14sin⁡3π14sin⁡5π14sin⁡7π14sin⁡9π14sin⁡11π14sin⁡13π14=

=sin⁡π14sin⁡3π14sin⁡5π142=−cos⁡π7cos⁡2π7cos⁡4π72=sin⁡8π78sin⁡π72=164

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$If \cos \frac{π}{7} \cos \frac{2 π}{7} \cos \frac{4 π}{7} = \frac{\sin \frac{8 π}{7}}{8 \sin \frac{π}{7}} then, \sin \frac{π}{14} \sin \frac{3 π}{14} \sin \frac{5 π}{14} \sin \frac{7 π}{14} \sin \frac{9 π}{14} \sin \frac{11 π}{14} \sin \frac{13 π}{14} =$

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$\frac{1}{16}$

$\frac{1}{32}$

$\frac{1}{64}$

$\frac{1}{128}$

answer is C.

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Detailed Solution

$\begin{array}{l} = {[\sin \frac{π}{14} \sin \frac{3 π}{14} \sin \frac{5 π}{14}]}^{2} \\ = {[- \cos \frac{π}{7} \cos \frac{2 π}{7} \cos \frac{4 π}{7}]}^{2} \\ = {[\frac{\sin \frac{8 π}{7}}{8 \sin \frac{π}{7}}]}^{2} \\ = \frac{1}{64} \end{array}$