| Direction: If \[\cos \,\frac{\pi }{7},\,\,\cos \,\frac{3\pi }{7}\]and \[\cos \,\frac{5\pi }{7}\] are the roots of the equations \[8{{x}^{3}}-4{{x}^{2}}-4x+1=0\] then, |
A) \[\frac{1}{4}\]
B) \[\frac{1}{8}\]
C) \[\frac{\sqrt{7}}{4}\]
D) \[\frac{\sqrt{7}}{8}\]
Correct Answer: B
Solution :
From Eq. (i), \[8{{x}^{3}}-4{{x}^{2}}-4x+1\] \[=8\,\left( x-\cos \,\frac{\pi }{7} \right)\,\left( x-\cos \,\frac{3\pi }{7} \right)\,\left( x-\cos \,\frac{5\pi }{7} \right)\] Put \[x=1,\] then \[1=8\,\left( 1-\cos \,\frac{\pi }{7} \right)\,\left( 1-\cos \,\frac{3\pi }{7} \right)\,\left( 1-\cos \,\frac{5\pi }{7} \right)\] \[\therefore \] \[{{\sin }^{2}}\,\frac{\pi }{14}{{\sin }^{2}}\frac{3\pi }{14}{{\sin }^{2}}\frac{5\pi }{14}=\frac{1}{64}\] \[\Rightarrow \] \[\sin \,\frac{\pi }{14}\sin \,\frac{3\pi }{14}\,\sin \,\frac{5\pi }{14}=\frac{1}{8}\]
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