A) \[\frac{1}{4}\]
B) \[\frac{1}{6}\]
C) \[\frac{1}{8}\]
D) \[\frac{1}{2}\]
Correct Answer: C
Solution :
\[\left[ 1+\cos \frac{\pi }{8} \right]\left[ 1+\cos \frac{3\pi }{8} \right]\left[ 1+\cos \frac{5\pi }{8} \right]\] \[\left[ 1+\cos \frac{7\pi }{8} \right]\] \[=\left[ 1+\cos \frac{\pi }{8} \right]\left[ 1+\cos \left( \frac{\pi }{2}-\frac{\pi }{8} \right) \right]\] \[\left[ 1+\cos \left( \frac{\pi }{2}+\frac{\pi }{8} \right) \right]\left[ 1+\cos \left( \pi -\frac{\pi }{8} \right) \right]\] \[=\left[ 1+\cos \frac{\pi }{8} \right]\left[ 1+\sin \frac{\pi }{8} \right]\left[ 1-\sin \frac{\pi }{8} \right]\] \[\left[ 1-\cos \frac{\pi }{8} \right]\] \[=\left[ 1-{{\cos }^{2}}\frac{\pi }{8} \right]\left[ 1-{{\sin }^{2}}\frac{\pi }{8} \right]\] \[={{\sin }^{2}}\frac{\pi }{8}.{{\cos }^{2}}\frac{\pi }{8}\] \[=\frac{1}{4}{{\left( 2\sin \frac{\pi }{8}\cos \frac{\pi }{8} \right)}^{2}}\] \[=\frac{1}{4}{{\left( \sin \frac{\pi }{4} \right)}^{2}}\] \[=\frac{1}{4}.\frac{1}{2}=\frac{1}{8}\]You need to login to perform this action.
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