# Finding a complex Taylor series

1. Apr 20, 2012

### dochalladay32

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

Since cosh(x) = cos(i*x), I can rewrite my original function as

cos(cos θ)*cos(i*sin θ).

3. 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)?

2. Apr 20, 2012

### Dick

No quirky identity can make cos(2nθ)=cos(nθ), because it's not true. I think you are right and there is a typo in the solution.

3. Apr 21, 2012

### dochalladay32

That's what I thought. Thanks for clarifying. I'll point it out on Monday in class and hopefully I'm not the only one that noticed that.

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