Simplification of answer involving Cosine

  • Thread starter zack7
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  • #1
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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
 

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  • #2
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[itex]\cos(x\pi)=(-1)^x[/itex] if [itex]x\in\mathbb{Z}[/itex]
Does that help?
 
  • #3
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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]
 
  • #4
haruspex
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You may be looking for something more complicated than necessary.
What I don't get is how to I simplify from
[itex]\frac{-(cos(n*π+π)}{(n+1)}[/itex]+[itex]\frac{-(cos(n*π-π)}{(n-1)}[/itex]
to
[itex]\frac{(cos(n*π)}{(n+1)}[/itex]-[itex]\frac{(cos(n*π)}{(n-1)}[/itex]
How would you simplify cos(θ+π) and cos(θ-π)?
to
[itex]\frac{-(2 cos(π n)}{(n^2-1)}[/itex]
How would combine a/(n+1) - a/(n-1) into a single fraction?
 
  • #5
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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?
Yup got it, totally forgot the trig identity

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

Thank you
 
  • #6
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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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