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Trigonometry and method of difference

  1. Nov 3, 2011 #1
    1. The problem statement, all variables and given/known data

    Given that 1+2cosx+2cos2x+2cos3x + .................+2cosnx = sin(x+1/2)x cosec(x/2)
    Prove that 1-2cosx+2cos2x-2cos3x+.................+2cos8x = cos(17/2)x sec(x/2)


    2. Relevant equations



    3. The attempt at a solution

    1+2cosx+2cos2x+2cos3x + .................+2cos8x = sin(8+1/2)x cosec(x/2)

    4cosx+4cos3x+4cos5x........ = ?

    I don't know how to continue. Or is there another way?
     
  2. jcsd
  3. Nov 4, 2011 #2

    NascentOxygen

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    Staff: Mentor

    1+2cosx+2cos2x+2cos3x + .................+ 2cos8x = ... eqn(1)
    Substitute 2x for x, and you can say
    1 + 2cos2x + 2 cos4x + 2cos6x + 2cos8x = ... eqn(2)

    If you subtract double eqn(2) from eqn(1) you'll as good as have the left hand side. I don't know whether this will get you anywhere.

    Also, fix up the x that should be n in your first line.
     
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