A) \[\frac{1}{2}\]
B) \[\cos \frac{\pi }{8}\]
C) \[\frac{1}{8}\]
D) \[\frac{1+\sqrt{2}}{2\sqrt{2}}\]
Correct Answer: C
Solution :
\[\left( 1+\cos \frac{\pi }{8} \right)\left( 1+\cos \frac{3\pi }{8} \right)\] \[\left( 1+\cos \frac{5\pi }{8} \right)\left( 1+\cos \frac{7\pi }{8} \right)\] \[=\left( 1+\cos \frac{\pi }{8} \right)\left( 1+\cos \frac{3\pi }{8} \right)\] \[\left( 1-\cos \frac{3\pi }{8} \right)\times \left( 1-\cos \frac{\pi }{8} \right)\] \[\left[ \begin{align} & \because \cos \frac{5\pi }{8}=\cos \left( \pi -\frac{3\pi }{8} \right)=-\cos 3\frac{\pi }{8} \\ & and\,\,\cos \frac{7\pi }{8}=\cos \left( \pi -\frac{\pi }{8} \right)=-\cos \frac{\pi }{8} \\ \end{align} \right]\] \[=\left( 1-{{\cos }^{2}}\frac{\pi }{8} \right)\left( 1-{{\cos }^{2}}\frac{3\pi }{8} \right)\] \[={{\sin }^{2}}\frac{\pi }{8}\cdot {{\sin }^{2}}\frac{3\pi }{8}=\frac{1}{4}{{\left[ 2\sin \frac{\pi }{8}\sin \frac{3\pi }{8} \right]}^{2}}\] \[=\frac{1}{4}{{\left[ \cos \frac{\pi }{4}-\cos \frac{\pi }{2} \right]}^{2}}=\frac{1}{4}{{\left[ \frac{1}{\sqrt{2}}-0 \right]}^{2}}=\frac{1}{4}\times \frac{1}{2}=\frac{1}{8}\]
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