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sin^(4)frac(pi)(8) + sin^(4) frac(2pi)8 ...

`sin^(4)frac(pi)(8) + sin^(4) frac(2pi)8 + sin^(4) frac(3pi)8+sin^(4) frac(4pi)(8)`
`+sin^(4)frac(5pi)(8) + sin^(4) frac(6pi)8 + sin^(4) frac(7pi)(8)=`

A

A `3/2`

B

B `5/2`

C

C 3

D

D `7/2`

Text Solution

Verified by Experts

The correct Answer is:
C
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Show that sin ^(4). (pi)/(8)+ sin ^(4) .(3pi)/(8) sin ^(4). (5pi)/(8)+sin ^(4). (7pi)/(8) =(3)/(2)

(ii) Find the value of sin^(2). (pi)/(4) + sin^(2). (3pi)/(4) + sin^(2). (5pi)/(4) + sin^(2). (7pi)/(4) .

Knowledge Check

  • "sin"^(4)(pi)/(16) + "sin"^(4) (3pi)/(16) + "sin"^(4) (5pi)/(16) + "sin"^(4) (7pi)/(16)=

    A
    1
    B
    `1//2`
    C
    `3//2`
    D
    2

  • "sin"^(4)pi/8" +sin"^(4)(3pi)/8" +sin"^(4)(5pi)/8" +sin"^(4)(7pi)/8=

    A
    `1/2`
    B
    `3/2`
    C
    `3/4`
    D
    `5/4`

  • "sin"^(2)pi/8" + sin"^(2)(3pi)/8" + sin"^(2)(5pi)/8" + sin"^(2)(7pi)/8=

    A
    1
    B
    2
    C
    4
    D
    6

    • Similar Questions

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    sin^(2)""(pi)/(18)+sin^(2)""(2pi)/(18)+sin^(2)""(4pi)/(18)+sin^(2)""(8pi)/(18)+sin^(2)""(7pi)/(18)+sin^(2)""(5pi)/(18)=

    cos((pi)/(4)+A) cos((pi)/(4) - B) + sin ((pi)/(4) +A) * sin((pi)/4-B)=

    cos^(2)""((pi)/(4) +x ) - sin^(2) ""((pi)/4- x ) =

    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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