If $cos \frac{2 π}{7}, cos \frac{4 π}{7}, cos \frac{6 π}{7}$ are roots of the equation $8 x^{3} + 4 x^{2} - 4 x - 1 = 0$ . The value of $cos \frac{π}{7} cos \frac{2 π}{7} cos \frac{3 π}{7}$ is equal to

\frac{1}{64}

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\frac{\sqrt{7}}{8}

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\frac{7}{64}

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\frac{1}{8}

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Solution

The correct option is C $\frac{1}{8}$
As $c o s \frac{2 π}{7}, c o s \frac{4 π}{7}, c o s \frac{6 π}{7}$ roots of $8 x^{3} + 4 x^{2} - 4 x - 1 = 0$
Therefore
$8 (x - c o s \frac{2 π}{7}) (x - c o s \frac{4 π}{7}) (x - c o s \frac{6 π}{7}) = 8 x^{3} + 4 x^{2} - 4 x - 1$
For $x = - 1$
$8 (- 1 - c o s \frac{2 π}{7}) (- 1 - c o s \frac{4 π}{7}) (- 1 - c o s \frac{6 π}{7}) = - 1$

$\Rightarrow 8 (- 1 + 2 {c o s}^{2} \frac{π}{7} + 1) (- 1 + 2 {c o s}^{2} \frac{2 π}{7} + 1) (- 1 + 2 {c o s}^{2} \frac{3 π}{7} + 1) = 1$

$\Rightarrow {c o s}^{2} \frac{π}{7} {c o s}^{2} \frac{2 π}{7} {c o s}^{2} \frac{3 π}{7} = \frac{1}{64} \Rightarrow c o s \frac{π}{7} c o s \frac{2 π}{7} c o s \frac{3 π}{7} = \frac{1}{8}$

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Similar questions

Q. If

cos \frac{π}{7}, cos \frac{3 π}{7}, cos \frac{5 π}{7}

are the roots of the equation

8 x^{3} - 4 x^{2} - 4 x + 1 = 0

The value of

sec \frac{π}{7} + sec \frac{3 π}{7} + sec \frac{5 π}{7} =

$c o s \frac{π}{7}$ . $c o s \frac{3 π}{7}$ . $c o s \frac{5 π}{7}$ are the roots of the equation $8 x^{3} - 4 x^{2} - 4 x + 1 = 0$ . Then the value of $s e c \frac{π}{7} + s e c \frac{3 π}{7} + s e c \frac{5 π}{7}$ is

cos \frac{2 π}{7} + cos \frac{4 π}{7} + cos \frac{6 π}{7} = - \frac{1}{2}

and

cos \frac{2 π}{7} cos \frac{4 π}{7} cos \frac{6 π}{7} = \frac{1}{8}

. Then the numerical value of

{c o s e c}^{2} \frac{π}{7} + {c o s e c}^{2} \frac{2 π}{7} + {c o s e c}^{2} \frac{3 π}{7}

Q. The value of

cos 0 + cos \frac{π}{7} + cos \frac{2 π}{7} + cos \frac{3 π}{7} + cos \frac{4 π}{7} + cos \frac{5 π}{7} + cos \frac{6 π}{7} =

1 + c o s \frac{2 π}{7} + c o s \frac{4 π}{7} + c o s \frac{6 π}{7}

is equal to

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If cos2π7,cos4π7,cos6π7 are roots of the equation 8x3+4x2−4x−1=0. The value of cosπ7cos2π7cos3π7 is equal to

If $cos \frac{2 π}{7}, cos \frac{4 π}{7}, cos \frac{6 π}{7}$ are roots of the equation $8 x^{3} + 4 x^{2} - 4 x - 1 = 0$ . The value of $cos \frac{π}{7} cos \frac{2 π}{7} cos \frac{3 π}{7}$ is equal to