If the value of sin⁡π14sin⁡3π14sin⁡5π14sin⁡7π14sin⁡9π14sin⁡11π14sin⁡13π14 is equal to k2 then k is equal to

The given expression can be written assin⁡π14sin⁡3π14sin⁡5π14sin⁡π2sin⁡π−5π14sin⁡π−3π14sin⁡π−π14=k2where k=sin⁡π14sin⁡3π14sin⁡5π14=cos⁡π2−π14cos⁡π2−3π14cos⁡π2−5π14=cos⁡3π7cos⁡2π7cos⁡π7 =cos⁡3π7cos⁡2π7cos⁡π7=12sin⁡π7×2sin⁡π7cos⁡π7cos⁡2π7cos⁡4π7=14sin⁡π7sin⁡4π7cos⁡4π7=18sin⁡π7×sin⁡8π7=−18

If the value of sin⁡π14sin⁡3π14sin⁡5π14sin⁡7π14sin⁡9π14sin⁡11π14sin⁡13π14 is equal to k2 then k is equal to

The given expression can be written assin⁡π14sin⁡3π14sin⁡5π14sin⁡π2sin⁡π−5π14sin⁡π−3π14sin⁡π−π14=k2where k=sin⁡π14sin⁡3π14sin⁡5π14=cos⁡π2−π14cos⁡π2−3π14cos⁡π2−5π14=cos⁡3π7cos⁡2π7cos⁡π7 =cos⁡3π7cos⁡2π7cos⁡π7=12sin⁡π7×2sin⁡π7cos⁡π7cos⁡2π7cos⁡4π7=14sin⁡π7sin⁡4π7cos⁡4π7=18sin⁡π7×sin⁡8π7=−18

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If the value of
$\sin \frac{π}{14} \sin \frac{3 π}{14} \sin \frac{5 π}{14} \sin \frac{7 π}{14} \sin \frac{9 π}{14} \sin \frac{11 π}{14} \sin \frac{13 π}{14}$ is equal to $k^{2}$ then k is equal to

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Detailed Solution

The given expression can be written as

$\begin{array}{r} \sin \frac{π}{14} \sin \frac{3 π}{14} \sin \frac{5 π}{14} \sin (\frac{π}{2}) \sin (π - \frac{5 π}{14}) \\ \sin (π - \frac{3 π}{14}) \sin (π - \frac{π}{14}) = k^{2} \end{array}$

where

$\begin{array}{l} k = \sin \frac{π}{14} \sin \frac{3 π}{14} \sin \frac{5 π}{14} \\ = \cos (\frac{π}{2} - \frac{π}{14}) \cos (\frac{π}{2} - \frac{3 π}{14}) \cos (\frac{π}{2} - \frac{5 π}{14}) \\ = \cos \frac{3 π}{7} \cos \frac{2 π}{7} \cos \frac{π}{7} \end{array}$

$\begin{array}{l} = \cos \frac{3 π}{7} \cos \frac{2 π}{7} \cos \frac{π}{7} \\ = \frac{1}{2 \sin \frac{π}{7}} \times 2 \sin \frac{π}{7} \cos \frac{π}{7} \cos \frac{2 π}{7} \cos \frac{4 π}{7} \\ = \frac{1}{4 \sin \frac{π}{7}} \sin \frac{4 π}{7} \cos \frac{4 π}{7} \end{array}$