2 sin left(frac{pi}{22}right) sin left(frac{3 | vedclass

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(B) Let $S = 2 \sin \frac{\pi}{22} \sin \frac{3 \pi}{22} \sin \frac{5 \pi}{22} \sin \frac{7 \pi}{22} \sin \frac{9 \pi}{22}$.
Using $\sin 	heta = \cos

Vedclass · Accepted Answer

$\frac{1}{16}$

Vedclass · Answer

(B) Let $S = 2 \sin \frac{\pi}{22} \sin \frac{3 \pi}{22} \sin \frac{5 \pi}{22} \sin \frac{7 \pi}{22} \sin \frac{9 \pi}{22}$.
Using $\sin 	heta = \cos \left(\frac{\pi}{2} - 	hetaight)$,we have:
$\sin \frac{\pi}{22} = \cos \frac{10 \pi}{22} = \cos \frac{5 \pi}{11}$
$\sin \frac{3 \pi}{22} = \cos \frac{8 \pi}{22} = \cos \frac{4 \pi}{11}$
$\sin \frac{5 \pi}{22} = \cos \frac{6 \pi}{22} = \cos \frac{3 \pi}{11}$
$\sin \frac{7 \pi}{22} = \cos \frac{4 \pi}{22} = \cos \frac{2 \pi}{11}$
$\sin \frac{9 \pi}{22} = \cos \frac{2 \pi}{22} = \cos \frac{\pi}{11}$
Thus,$S = 2 \cos \frac{\pi}{11} \cos \frac{2 \pi}{11} \cos \frac{3 \pi}{11} \cos \frac{4 \pi}{11} \cos \frac{5 \pi}{11}$.
Using the formula $\prod_{k=1}^{n} \cos \frac{k \pi}{2n+1} = \frac{1}{2^n}$,for $n=5$,we have $\prod_{k=1}^{5} \cos \frac{k \pi}{11} = \frac{1}{2^5} = \frac{1}{32}$.
Therefore,$S = 2 	imes \frac{1}{32} = \frac{1}{16}$.

$2 \sin \left(\frac{\pi}{22}\right) \sin \left(\frac{3 \pi}{22}\right) \sin \left(\frac{5 \pi}{22}\right) \sin \left(\frac{7 \pi}{22}\right) \sin \left(\frac{9 \pi}{22}\right)$ is equal to:

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