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