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Trig id's and fourier

  1. Jan 11, 2004 #1


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    hi, first off just have to say these boards kick serious ass!!! Too much interesting posts on here, really like the Mechanical eng board since thats what im doing at uni :)

    Right, to the problem,

    I do the integral and get the below answer,

    Bn = 1/pi [-Cos nt/n] between pi/2 and 0
    Bn = 1/pi [[-Cos n(pi/2)/n]-1/n]

    but then my notes skip from this to another line,

    Bn = 1/npi [1-Cos nt]

    The n gets taken out and 1 comes from Cos n0 but why do the signs change?
  2. jcsd
  3. Jan 11, 2004 #2


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    LOL just looked at it again, and i forgot that there's a minus infront of the cos nt/n which makes

    Bn = 1/npi [(-cos nt)-(-1)]

    which answers my question :)
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