Home
Class 12
MATHS
"cos " pi/7 - "cos " (2pi)/7 + "cos" (3p...

`"cos " pi/7 - "cos " (2pi)/7 + "cos" (3pi)/7 - "cos" (4pi)/7 + "cos "(5pi)/7 - "cos"(6pi)/7` =

A

0

B

`3/2`

C

`3/4`

D

1

Text Solution

Verified by Experts

The correct Answer is:
D
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • EAMCET-2019 (TS) SHIFT-1

    SIA PUBLICATION|Exercise PHYSICS|40 Videos

  • EAMCET-2019 (TS) SHIFT-1

    SIA PUBLICATION|Exercise CHEMISTRY|10 Videos

  • EAMCET QUESTION PAPER 2017

    SIA PUBLICATION|Exercise CHEMISTRY|6 Videos

  • FUNCTIONS AND MATHEMATICAL INDUCTION

    SIA PUBLICATION|Exercise Problems|58 Videos

Similar Questions

Explore conceptually related problems

{:(,"Column-I",,"Column-II"),((A),cos 20^(0) + cos 80^(0) - sqrt(3) "cos" 50^(0),(p),-1),((B),cos 0^(0) + "cos" (pi)/(7) + "cos" (2pi)/(7) + "cos" (4pi)/(7) + "cos" (5pi)/(7) + "cos" (6 pi)/(7),(q),-(3)/(4)),((C),cos 20^(0) + cos 40^(0) + cos 60^(0) - 4 cos 10^(0) "cos" 20^(0)"cos" 30^(0),(r),1),((D),"cos" 20^(0) cos 100^(0) + cos 100^(0) "cos" 140^(0) - cos 140^(0) "cos" 200^(0),(s),0):}

Suppose "cos" (2pi)/(7) + "cos" (4pi)/(7) + "cos" (6pi)/(7) = - (1)/(2) and "cos" (2pi)/(7) "cos"(4pi)/(7) "cos" (6pi)/(7) = - (1)/(8) , then the numerical value of "cosec"^(2) (pi)/(7) + "cosec"^(2)(2pi)/(7) + "cosec"^(2) (3pi)/(7) must be

Knowledge Check

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

    A
    `(1)/(2)`
    B
    `-(1)/(2)`
    C
    0
    D
    1

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

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

  • If "cos"(pi)/(15) "cos"(2pi)/(15) "cos"(4pi)/(15) "cos"(5pi)/(15) "cos"(7pi)/(15) "cos"(30pi)/(15) =x, " then " (1)/(8x) =

    A
    4
    B
    `(1)/(4)`
    C
    8
    D
    `(4)/(3)`

    • Similar Questions

    Explore conceptually related problems

    Suppose cos""(2pi)/7+ cos""(4pi)/7 + cos""(6pi)/7 =-1/2 and cos""(2pi)/7cos""(4pi)/7cos""(6pi)/7=1/8 , then the numerical value of "cosec"^(2)(pi)/7 + "cosec"^(2)(2pi)/7+ "cosec"^2(3pi)/7 must be

    Prove that 2 "cos"(pi)/13"cos"(9pi)/13+"cos"(3pi)/13+"cos"(5pi)/13=0

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

    sin "" (pi)/(7) sin "" (2pi)/(7) sin "" (4pi )/(7)=

    cos""(2pi)/(15) cos ""(4pi)/(15)""cos(8pi)/(15) cos""(16pi)/(15)=

    Home
    Profile