| Directions: Question based on the following paragraph. |
| If\[\cos \frac{\pi }{7},\cos \frac{3\pi }{7},\cos \frac{5\pi }{7}.\]are the roots of the equation \[8{{x}^{3}}-4{{x}^{b}}-4x+1=0\]. |
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, we get \[1=8\left( 1-\cos \frac{\pi }{7} \right)\left( 1-\cos \frac{3\pi }{7} \right)\left( 1-\cos \frac{5\pi }{7} \right)\] \[\Rightarrow \]\[1=8\left( 2{{\sin }^{2}}\frac{\pi }{17} \right)\left( 2{{\sin }^{2}}\frac{3\pi }{14} \right)\left( 2{{\sin }^{2}}\frac{5\pi }{14} \right)\] \[\therefore \]\[\sin \left( \frac{\pi }{17} \right)\sin \left( \frac{3\pi }{14} \right)\sin \left( \frac{5\pi }{14} \right)=\frac{1}{8}\]
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