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Simplification of answer involving Cosine

  1. Nov 23, 2012 #1
    I have manged to get my answer down to the first line in the picture but I have tried all ways and can't seem to simplify it to the second line.

    Thank you
     

    Attached Files:

  2. jcsd
  3. Nov 23, 2012 #2
    [itex]\cos(x\pi)=(-1)^x[/itex] if [itex]x\in\mathbb{Z}[/itex]
    Does that help?
     
  4. Nov 23, 2012 #3
    What I don't get is how to I simplify from
    [itex]\frac{-(cos(n*pi+pi)}{(n+1)}[/itex]+[itex]\frac{-(cos(n*pi-pi)}{(n-1)}[/itex]

    to
    [itex]\frac{(cos(n*pi)}{(n+1)}[/itex]-[itex]\frac{(cos(n*pi)}{(n-1)}[/itex]

    to
    [itex]\frac{-(2 cos(pi n)}{(n^2-1)}[/itex]
     
  5. Nov 23, 2012 #4

    haruspex

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    You may be looking for something more complicated than necessary.
    How would you simplify cos(θ+π) and cos(θ-π)?
    How would combine a/(n+1) - a/(n-1) into a single fraction?
     
  6. Nov 23, 2012 #5
    Yup got it, totally forgot the trig identity

    cos(A+ B)= cos(A)cos(B)- sin(A)sin(B)

    Thank you
     
  7. Nov 24, 2012 #6
    You do not actually need that identity for this, the rather simple identities [itex]\cos(\pi+\theta)=-\cos(\theta)[/itex] and [itex]\cos(\theta-\pi)=\cos(\pi-\theta)=-\cos(\theta)[/itex] are enough.
     
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