$2 \sin (\frac{π}{22}) \sin (\frac{3 π}{22}) \sin (\frac{5 π}{22}) \sin (\frac{7 π}{22}) \sin (\frac{9 π}{22})$ is equal to

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$\frac{1}{16}$

$\frac{3}{16}$

$\frac{1}{32}$

$\frac{9}{32}$

answer is B.

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Detailed Solution

$2 \sin (\frac{π}{22}) \sin (\frac{3 π}{22}) \sin (\frac{5 π}{22}) \sin (\frac{7 π}{22}) \sin (\frac{9 π}{22})$
$\begin{aligned} = 2 \sin (\frac{11 π - 10 π}{22}) \sin (\frac{11 π - 8 π}{22}) \sin (\frac{11 π - 6 π}{22}) \\ \sin (\frac{11 π - 4 π}{22}) \sin (\frac{11 π - 2 π}{22}) \end{aligned}$
$\begin{array}{l} = 2 \cos \frac{π}{11} \cos \frac{2 π}{11} \cos \frac{3 π}{11} \cos \frac{4 π}{11} \cos \frac{5 π}{11} \\ = 2 \cos \frac{π}{11} \cos \frac{2 π}{11} \cos (π - \frac{8 π}{11}) \cos \frac{4 π}{11} \cos (π + \frac{5 π}{11}) \\ = 2 \cos \frac{π}{11} \cos \frac{2 π}{11} \cos \frac{4 π}{11} \cos \frac{8 π}{11} \cos \frac{16 π}{11} \\ \cos θ \cos 2 θ \cos 2^{2} θ . . . . . \cos 2^{n - 1} θ = \frac{\sin 2^{n} θ}{2^{n} \sin θ} \\ = \frac{2 \sin \frac{32 π}{11}}{2^{5} \sin \frac{π}{11}} \end{array}$
$= \frac{1}{16}$