A) \[\frac{1}{16}\]
B) \[\frac{\sqrt{2}}{16}\]
C) \[\frac{1}{8}\]
D) \[\frac{\sqrt{2}}{8}\]
Correct Answer: B
Solution :
We have \[\sin \frac{\pi }{16}.\sin \frac{3\pi }{16}.\sin \frac{5\pi }{16}.\sin \frac{7\pi }{16}\] \[=\frac{1}{4}\left[ 2\sin \frac{\pi }{16}\sin \frac{3\pi }{16}.2\sin \frac{5\pi }{16}\sin \frac{7\pi }{16} \right]\] \[=\frac{1}{4}\left[ \left( \cos \frac{\pi }{8}-\cos \frac{\pi }{4} \right)\left( \cos \frac{\pi }{8}-\cos \frac{3\pi }{4} \right) \right]\] \[=\frac{1}{4}\left[ \left( \cos \frac{\pi }{8}-\frac{1}{\sqrt{2}} \right)\left( \cos \frac{\pi }{8}+\frac{1}{\sqrt{2}} \right) \right]\] \[=\frac{1}{4}\left[ \left( {{\cos }^{2}}\frac{\pi }{8}-\frac{1}{2} \right) \right]=\frac{1}{8}\left[ 2{{\cos }^{2}}\frac{\pi }{8}-1 \right]\] \[=\frac{1}{8}\left[ \cos \frac{\pi }{4} \right]=\frac{1}{8}\times \frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{16}\].
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