sin⁡π14⋅sin⁡3π14⋅sin⁡5π14⋅sin⁡7π14⋅sin⁡9π14⋅sin⁡11π14⋅sin⁡13π14=

=sin⁡π14sin⁡3π14sin⁡5π14sin⁡π2sin⁡π14sin⁡3π14sin⁡5π14Write sin⁡π14sin⁡3π14sin⁡5π142=cos⁡π7cos⁡2π7cos⁡4π72and use the formulacos⁡θcos⁡2θcos⁡4θ….cos⁡2(n−1)θ=2sin⁡2n−1θ2nsin⁡θ=sin⁡23θ23sin⁡θ2=18sin⁡8π7sin⁡π72=164(−1)2=164

sin⁡π14⋅sin⁡3π14⋅sin⁡5π14⋅sin⁡7π14⋅sin⁡9π14⋅sin⁡11π14⋅sin⁡13π14=

=sin⁡π14sin⁡3π14sin⁡5π14sin⁡π2sin⁡π14sin⁡3π14sin⁡5π14Write sin⁡π14sin⁡3π14sin⁡5π142=cos⁡π7cos⁡2π7cos⁡4π72and use the formulacos⁡θcos⁡2θcos⁡4θ….cos⁡2(n−1)θ=2sin⁡2n−1θ2nsin⁡θ=sin⁡23θ23sin⁡θ2=18sin⁡8π7sin⁡π72=164(−1)2=164

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$\sin \frac{π}{14} \cdot \sin \frac{3 π}{14} \cdot \sin \frac{5 π}{14} \cdot \sin \frac{7 π}{14} \cdot \sin \frac{9 π}{14} \cdot \sin \frac{11 π}{14} \cdot \sin \frac{13 π}{14} =$

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$\frac{1}{64}$

$\frac{3}{64}$

$\frac{5}{64}$

$\frac{7}{64}$

answer is A.

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Detailed Solution

$= \sin \frac{π}{14} \sin \frac{3 π}{14} \sin \frac{5 π}{14} \sin \frac{π}{2} \sin \frac{π}{14} \sin \frac{3 π}{14} \sin \frac{5 π}{14}$
Write ${(\sin \frac{π}{14} \sin \frac{3 π}{14} \sin \frac{5 π}{14})}^{2} = {(\cos \frac{π}{7} \cos \frac{2 π}{7} \cos \frac{4 π}{7})}^{2}$
and use the formula
$\begin{array}{l} \cos θcos 2 θcos 4 θ \dots . \cos 2^{(n - 1)} θ = \frac{2 \sin 2^{n - 1} θ}{2^{n} \sin θ} \\ = {(\frac{\sin 2^{3} θ}{2^{3} \sin θ})}^{2} = {(\frac{1}{8} \frac{\sin 8 \frac{π}{7}}{\sin \frac{π}{7}})}^{2} = \frac{1}{64} (- 1)^{2} = \frac{1}{64} \end{array}$