Help with a trig to sum identity.

  • Thread starter Gallani
  • Start date
  • #1
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I have been working on showing the equality between

N=0 to ∞ Ʃ cos(2nθ)(-1)^n/(2n)! = cos(cos(θ))cosh(sin(θ))

I started by using the standard series for cosine and putting cos(2nθ) in for the x term.

I did the same for cosh(sin(θ)). I manipulated the forms every way I could think but it never looked anything like the answer I seek.
 

Answers and Replies

  • #2
392
17
I have been working on showing the equality between

N=0 to ∞ Ʃ cos(2nθ)(-1)^n/(2n)! = cos(cos(θ))cosh(sin(θ))

I started by using the standard series for cosine and putting cos(2nθ) in for the x term.

I did the same for cosh(sin(θ)). I manipulated the forms every way I could think but it never looked anything like the answer I seek.

You don't want to put ##\cos{(2n \theta)}## into the formula, you want ##\cos{(\theta)}##. Maybe that's where you went wrong?
 

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