If the roots of the equation 8x3-4x2-4x-1-0 are cos-7-cos3-7-cos5-7- then show that -7-3-7-5-7-4 -

As nbsp;cosπ nbsp;/ 7 nbsp; ,cos 3π nbsp;/ 7 nbsp; ,cos 5π nbsp;/ 7 nbsp; are roots of 8x ^ 3 4x ^ 2 4x+1=0 Then nbsp; 1 /cosπ ...

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Byju's Answer

Standard XII

Mathematics

Circular Measurement of Angle

If the roots ...

Question

If the roots of the equation $8 x^{3} - 4 x^{2} - 4 x + 1 = 0$ are $cos \frac{π}{7}, cos \frac{3 π}{7}, cos \frac{5 π}{7}$ , then show that $sec \frac{π}{7} + sec \frac{3 π}{7} + sec \frac{5 π}{7} = 4$ .

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Solution

As $cos \frac{π}{7}, cos \frac{3 π}{7}, cos \frac{5 π}{7}$ are roots of ${8 x}^{3} - {4 x}^{2} - 4 x + 1 = 0$
Then $\frac{1}{cos \frac{π}{7}}, \frac{1}{cos \frac{3 π}{7}}, \frac{1}{cos \frac{5 π}{7}}$ are roots of equation which we get by replacing $x \to \frac{1}{x}$
$∴ 8 {(\frac{1}{x})}^{3} - 4 {(\frac{1}{x})}^{2} - 4 (\frac{1}{x}) + 1 = 0 \Rightarrow 8 - 4 x - {4 x}^{2} + x^{3} = 0$
Hence sum of roots $= 4$
$∴ \frac{1}{cos \frac{π}{7}} + \frac{1}{cos \frac{3 π}{7}} + \frac{1}{cos \frac{5 π}{7}} = 4$
$\Rightarrow sec \frac{π}{7} + sec \frac{3 π}{7} + sec \frac{5 π}{7} = 4$

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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

$c o s \frac{π}{7}$ . $c o s \frac{3 π}{7}$ . $c o s \frac{5 π}{7}$ are the roots of the equation $8 x^{3} - 4 x^{2} - 4 x + 1 = 0$ . Then the value of $s i n \frac{π}{14} . s i n \frac{3 π}{14} . s i n \frac{5 π}{14}$ is

Q. If

cos \frac{2 π}{7}, cos \frac{4 π}{7}, cos \frac{6 π}{7}

are roots of the equation

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.The value of

cos \frac{π}{7} cos \frac{2 π}{7} cos \frac{3 π}{7}

is equal to

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If the roots of the equation 8x3−4x2−4x+1=0 are cosπ7,cos3π7,cos5π7, then show that secπ7+sec3π7+sec5π7=4 .

As cosπ7,cos3π7,cos5π7 are roots of 8x3−4x2−4x+1=0Then 1cosπ7,1cos3π7,1cos5π7 are roots of equation which we get by replacing x→1x∴8(1x)3−4(1x)2−4(1x)+1=0⇒8−4x−4x2+x3=0Hence sum of roots =4∴1cosπ7+1cos3π7+1cos5π7=4⇒secπ7+sec3π7+sec5π7=4

If the roots of the equation $8 x^{3} - 4 x^{2} - 4 x + 1 = 0$ are $cos \frac{π}{7}, cos \frac{3 π}{7}, cos \frac{5 π}{7}$ , then show that $sec \frac{π}{7} + sec \frac{3 π}{7} + sec \frac{5 π}{7} = 4$ .