If k = sin frac{pi}{18} cdot sin frac{5pi}{18 | vedclass

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(B) Given $k = \sin \frac{\pi}{18} \sin \frac{5\pi}{18} \sin \frac{7\pi}{18}$.Using the identity $\sin 	heta = \cos \left( \frac{\pi}{2} - 	heta i

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$\frac{1}{8}$

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(B) Given $k = \sin \frac{\pi}{18} \sin \frac{5\pi}{18} \sin \frac{7\pi}{18}$.Using the identity $\sin 	heta = \cos \left( \frac{\pi}{2} - 	heta ight)$:$k = \cos \left( \frac{\pi}{2} - \frac{\pi}{18} ight) \cos \left( \frac{\pi}{2} - \frac{5\pi}{18} ight) \cos \left( \frac{\pi}{2} - \frac{7\pi}{18} ight)$$k = \cos \frac{8\pi}{18} \cos \frac{4\pi}{18} \cos \frac{2\pi}{18} = \cos \frac{4\pi}{9} \cos \frac{2\pi}{9} \cos \frac{\pi}{9}$.Using the formula $\cos 	heta \cos 2	heta \cos 4	heta = \frac{\sin 8	heta}{8 \sin 	heta}$ where $	heta = \frac{\pi}{9}$:$k = \frac{\sin \left( 8 \cdot \frac{\pi}{9} ight)}{8 \sin \frac{\pi}{9}} = \frac{\sin \frac{8\pi}{9}}{8 \sin \frac{\pi}{9}}$.Since $\sin \frac{8\pi}{9} = \sin \left( \pi - \frac{\pi}{9} ight) = \sin \frac{\pi}{9}$:$k = \frac{\sin \frac{\pi}{9}}{8 \sin \frac{\pi}{9}} = \frac{1}{8}$.

If $k = \sin \frac{\pi}{18} \cdot \sin \frac{5\pi}{18} \cdot \sin \frac{7\pi}{18}$,then the numerical value of $k$ is

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