
Will B. answered 03/30/17
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Experienced & Knowledgeable Math Tutor - UGA Grad
So the easiest way to go about doing this would be to focus on the 2cos(θ). Since you are saying that this has to do with number theory/and or complex numbers I will assume we have our full book of tips and tricks at our disposal.
So what do we know about 2cos(θ)?
The trick here is to start with the basic property of eiθ
eiθ=cos(θ)+isin(θ) let this be Equation 1
similarly e-iθ=cos(θ)-isin(θ) let this be Equation 2
Now take Equation 1 + Equation 2
we get eiθ+e-iθ=2cos(θ)
This can then be generalized for n into 2cos(nθ)=eniθ+e-niθ
Now if we let x=eiθ we get 2cos(nθ)=xn+x-n which I do believe is what you wanted to show in the first place.
This equality can also be shown through induction, however it is not the prettiest thing in the world (but indeed doable).
Hope this helps!
Patrick L.
The second I am not sure how to type.
Evaluate ∑ with infinity on top and (n-0) on bottom. In front of sigma is [(cos(nθ))/(2^(n))], where cosθ = 1/5.
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03/30/17
Patrick L.
03/30/17