Find The Numerical Value Of (sin12x + 3sin10x + 3sin8x + sin6x – 1)

Cos x + cos2x = 1, then the numerical value of (sin12x + 3sin10x + 3sin8x + sin6x – 1) is

  1. 0
  2. 1
  3. -1
  4. 2
Anurag Mishra Professor Asked on 1st August 2015 in Maths.
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  • 1 Answer(s)
    Best answer

    Answer: (1) 0
    Explanation:-
    Cos X + cos2x = 1
    cos x = 1 – cos2x
    cos x = sin2x                                ……………………………. (1)
    Again, cos x + cos2x = 1
    cube on both side,
    (cos x + cos2x)3 = 13
    cos3x + (cos2x)3 + 3 cos x X cos2x(cos x + cos2x) = 1
    cos3x + cos6x + 3cos3x (cos x + cos2x) = 1
    cos3x + cos6x + 3cos4x + 3cos5 – 1 = 0
    Put the value of cos x = sin2x      [from equation (1)]
    (sin2x)3 + (sin2x)6 + 3(sin2x)4 + 3(sin2x)5 – 1 = 0
    sin6x + sin12x + 3sin8x + 3sin10x – 1 = 0
    sin12x + 3sin10x + 3sin8x + 3sin6x – 1 = 0
    Hence, the answer is (1) 0. 

    Anurag Mishra Professor Answered on 1st August 2015.
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