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