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Simplifying with trig identities

  1. May 15, 2015 #1
    1. The problem statement, all variables and given/known data

    Hi, I am currently working through a textbook, and the following simplification is given for an example question:

    upload_2015-5-15_11-48-40.png

    I can't seem to work out how they have moved from cos(pi+n*pi) to cos(pi)cos(n*pi) so easily? Is there a simple trick I have missed? I understand the identity that separates out the sine and cosine terms (-cos(a+b) = sin(a)sin(b)-cos(a)cos(b)) but I'm having very little luck in getting the textbooks answer from that.


    2. Relevant equations

    cos(a+b) = cos(a)cos(b)-sin(a)sin(b)
    cos(a-b) = cos(a)cos(b)+sin(a)sin(b)
    -cos(a+b) = sin(a)sin(b)-cos(a)cos(b)
    -cos(a-b) = -sin(a)sin(b)-cos(a)cos(b)

    3. The attempt at a solution
    applying the identity to the cosine/sine part of the equation ONLY (ignoring 2/(1-n^2) ) where a = pi :

    [itex]
    -\frac{\left(n + 1\right)\, \left(\cos\!\left(a\, n\right)\, \cos\!\left(a\right) + \sin\!\left(a\, n\right)\, \sin\!\left(a\right)\right) + \left(n - 1\right)\, \left(\cos\!\left(a\, n\right)\, \cos\!\left(a\right) - \sin\!\left(a\, n\right)\, \sin\!\left(n\right)\right)}{n^2 - 1}
    [/itex]

    Am I on the right track?

    Thanks
     
  2. jcsd
  3. May 15, 2015 #2

    jedishrfu

    Staff: Mentor

    Look at the cos(a+b) identity, what happens when you have sin(n*pi)?
     
  4. May 15, 2015 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    n is assumed to be an integer here, correct?
     
  5. May 15, 2015 #4
    jedishrfu... sorry not sure what you mean? Can you give a bit more detail?
    Hallsofivy...yes n is an integer
     
  6. May 15, 2015 #5
    oh! Thanks guys.... Completely couldn't see the wood for the trees :)

    sin(n*pi) = 0

    Doh!

    Cheers
     
  7. May 15, 2015 #6

    jedishrfu

    Staff: Mentor

    Sometimes the wood is fossilized. :-)
     
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