$c o s \frac{π}{7}$ . $c o s \frac{3 π}{7}$ . $c o s \frac{5 π}{7}$ are the roots of the equation $8 x^{3} - 4 x^{2} - 4 x + 1 = 0$ . Then the value of $s i n \frac{π}{14} . s i n \frac{3 π}{14} . s i n \frac{5 π}{14}$ is

$\frac{1}{4}$

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$\frac{1}{8}$

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$\frac{\sqrt{1}}{4}$

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$\frac{\sqrt{7}}{4}$

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Solution

The correct option is B
$\frac{1}{8}$

$8 x^{3} - 4 x^{2} - 4 x + 1$
$= 8 (x - c o s \frac{π}{7}) (x - c o s \frac{3 π}{7}) (x - c o s \frac{5 π}{7})$

Put x = 1, then

$1 = 8 (1 - c o s \frac{π}{7}) (1 - c o s \frac{3 π}{7}) (1 - c o s \frac{5 π}{7})$

$= 1 = 8 (2 s i n^{2} \frac{π}{14}) (2 s i n^{2} \frac{3 π}{14}) (2 s i n^{2} \frac{5 π}{14})$

$∴ s i n (\frac{π}{14}) (s i n \frac{3 π}{14}) (s i n \frac{5 π}{14}) = \frac{1}{8}$

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