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If 7 theta = (2n +1) pi, where n = 0 , ...

If `7 theta = (2n +1) pi, ` where `n = 0 , 1, 2, 3, 4,5, 6` then on the basis of above information, answer the following questions
The equation whose roots are `cos pi//7 , cos 3pi//7, cos 5pi//7` is

A

`8x^(3) + 4x^(2)+4x+1=0`

B

`8x^(3) - 4x^(2)-4x+1=0`

C

`8x^(3) - 4x^(2)-4x-1=0`

D

`8x^(3) + 4x^(2)+4x-1=0`

Text Solution

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The correct Answer is:
B
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Knowledge Check

  • If 7 theta = (2n +1) pi, where n = 0 , 1, 2, 3, 4,5, 6 then on the basis of above information, answer the following questions The value of sec pi//7 + sec 3pi//7 +sec 5pi//7 is

    A
    -3
    B
    4
    C
    `-4`
    D
    3

  • cos (pi/7) cos ((2pi)/(7) ) cos ( (4pi )/(7) ) =

    A
    `(-1)/(8)`
    B
    `1/8`
    C
    `-(3 sqrt3)/(8)`
    D
    1

  • If cos theta = cos 5pi//4 , then theta =

    A
    `2n pi pm (pi)/(4)`
    B
    `2n pi pm (3pi)/(4)`
    C
    `2n pi pm (5pi)/(4)`
    D
    `2n pi pm (7pi)/(4)`

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    The sum of all the solution is of the equation cos theta cos ((pi)/3+theta)cos ((pi)/3-theta)=1/4, theta in [0,6pi ]

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    "cos" (2pi)/(7)+"cos" (4pi)/(7)+"cos"(6pi)/(7)=

    To find the sum "sin"^(2) (2pi)/(7) + "sin"^(2) (4pi)/(7) + "sin"^(2) (8pi)/(7) , we follow the following method. Put 7 theta = 2 n pi , where n is any integer. Then sin 4 theta = sin (2n pi - 3 theta) = - sin 3 theta" "…(i) This means that sin theta takes the values 0. +- sin (2pi//7), +- sin (4pi//7), and +- sin (8pi//7) . From Eq. (i), we now get 2 sin 2theta cos 2 theta = 4 sin^(3) theta - 3 sin theta or 4 sin theta cos theta(1 - 2 sin^(2) theta) = (4 sin^(2) theta - 3) sin theta Rejecting the value sin theta = 0 , we get 4 cos theta (1-2 sin^(2) theta) = 4 sin^(2) theta - 3 or 16 cos^(2) theta (1-2sin^(2)theta)^(2) = (4 sin^(2) theta - 3)^(2) or 16(1-sin^(2)theta)(1-4 sin^(2)theta + 4 sin^(4) theta) = 16 sin^(4) theta - 24 sin^(2) theta + 9 or 64 sin^(6) theta - 112 sin^(4) theta - 56 sin^(2) theta - 7 = 0 , and this is cubic in sin^(2) theta with the roots sin^(2) ((2pi)/(7)),sin^(2)((4pi)/(7))and sin^(2) ((8pi)/(7)) The sum of these roots is "sin"^(2) (2pi)/(7) + "sin"^(2) (4pi)/(7) + "sin"^(2) (8pi)/(7) = (112)/(64) = (7)/(4) . The value of ("tan"^(2) (pi)/(7)+"tan"^(2)(2pi)/(7)+"tan"^(2)(3pi)/(7))xx("cot"^(2)(pi)/(7)+"cot"^(2) (2pi)/(7)+"cot"^(2) (3pi)/(7)) is

    To find the sum "sin"^(2) (2pi)/(7) + "sin"^(2) (4pi)/(7) + "sin"^(2) (8pi)/(7) , we follow the following method. Put 7 theta = 2 n pi , where n is any integer. Then sin 4 theta = sin (2n pi - 3 theta) = - sin 3 theta" "…(i) This means that sin theta takes the values 0. +- sin (2pi//7), +- sin (4pi//7), and +- sin (8pi//7) . From Eq. (i), we now get 2 sin 2theta cos 2 theta = 4 sin^(3) theta - 3 sin theta or 4 sin theta cos theta(1 - 2 sin^(2) theta) = (4 sin^(2) theta - 3) sin theta Rejecting the value sin theta = 0 , we get 4 cos theta (1-2 sin^(2) theta) = 4 sin^(2) theta - 3 or 16 cos^(2) theta (1-2sin^(2)theta)^(2) = (4 sin^(2) theta - 3)^(2) or 16(1-sin^(2)theta)(1-4 sin^(2)theta + 4 sin^(4) theta) = 16 sin^(4) theta - 24 sin^(2) theta + 9 or 64 sin^(6) theta - 112 sin^(4) theta - 56 sin^(2) theta - 7 = 0 , and this is cubic in sin^(2) theta with the roots sin^(2) ((2pi)/(7)),sin^(2)((4pi)/(7))and sin^(2) ((8pi)/(7)) The sum of these roots is "sin"^(2) (2pi)/(7) + "sin"^(2) (4pi)/(7) + "sin"^(2) (8pi)/(7) = (112)/(64) = (7)/(4) . The value of ("tan"^(2)(pi)/(7)+"tan"^(2)(2pi)/(7)+"tan"^(2) (3pi)/(7))/("cot"^(2)(pi)/(7)+"cot"^(2) (2pi)/(7) + "cot"^(2) (3pi)/(7)) is

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