Help with a trig to sum identity.

[Gallani]

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.

 

[scurty]

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?