- #1
dochalladay32
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Homework Statement
Not much has gotten me in this class, and I almost want to say this has to be a typo, but I want someone else to check it out first.
Homework question is that we need to show that
cos(cos θ)*cosh(sin θ) = Ʃ(-1)ncos(nθ)/(2n)! for n>=0
There is a similar one involving sin and sinh, but if the problem is sorted out here on this one, I can fix it for the other one.
Homework Equations
Since cosh(x) = cos(i*x), I can rewrite my original function as
cos(cos θ)*cos(i*sin θ).
The Attempt at a Solution
Using a product-to-sum trig identity, this eventually gets me to
1/2*[cos(exp(-iθ))+cos(exp(iθ))].
But because of the Taylor series for cosine, I end up with an exp(-2inθ) and exp(2inθ).
So I end up getting the series I expect, except I have a cos(2nθ) which does not match. Is there a division by 2 I'm missing somewhere or some quirky identity that makes it true that cos(2nθ)=cos(nθ)?