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15 November, 03:22

Prove:Sinx-2sin3x+sin5x=2sinx (cos4x-cos2x)

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Answers (2)
  1. 15 November, 03:29
    0
    A = sinx - sin3x,

    B = - sin3x + sin5x

    First A:

    The average of x and 3x is 2x, and they (x and 3x, that is) are each a distance of x from this average. That's fancy talk for:

    x = 2x-x,

    3x = 2x+x.

    So, A = sin (2x-x) - sin (2x+x)

    Using angle sum formulas:

    A = (sin2x cosx - cos2x sinx) - (sin2x cosx + cos2x sinx)

    A = - 2 cos2x sinx

    Similarly,

    B = - sin (4x-x) + sin (4x+x)

    = - (sin4x cosx - cos4x sinx) + (sin4x cosx + cos4x sinx)

    B = 2 cos4x sinx

    Now,

    sinx - 2sin3x + sin5x = A+B = - 2 cos2x sinx + 2 cos4x sinx

    = 2 sinx (cos4x - cos2x).
  2. 15 November, 03:50
    0
    Sinx-2sin3x+sin5x=2sinx (cos4x-cos2x)

    Step-by-step explanation:

    sinx - 2sin3x + sin5x = sinx - sin (3x) + sin (5x) - sin (3x)

    = 2· cos[ (x+3x) / 2] · sin[ (x-3x) / 2] + 2·cos[ (5x+3x) / 2]· sin[ (5x-3x) / 2]

    = 2· cos (2x) ·sin (-x) + 2· cos (4x) · sin (x)

    = - 2·cos (2x) ·sinx + 2· cos (4x) ·sinx

    = 2·sinx · [ cos (4x) - cos (2x) ]
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