TITLE: Prove: Fourier series of $e^{\cos x} \sin (\sin x)$ is $\sum_{n=0}^{\infty}\frac{\sin (nx)}{n!}$
QUESTION [9 upvotes]: I'd love your help with proving that the following series  $$\sum_{n=0}^{\infty}\frac{\sin (nx)}{n!}$$ is the Fourier series of $e^{\cos x}  \sin (\sin x)$.
I tried to find $\hat f(n)$ using integration by parts and I tried to use Taylor's series of $e^x$ in order to get $n!$ by I didn't reach to anything close of what I should.
Thanks a lot.

REPLY [18 votes]: I will elaborate Misty's answer: $e^{e^{ix}}=e^{\cos x + i \sin x}=e^{\cos x} (\cos (\sin x) +i \sin (\sin x ))$, and the second way of looking at $e^{e^{ix}}$ is:
$$ e^{e^{ix}}= \sum_{n=0}^{\infty} (e^{ix})^n/n!=\sum_{n=0}^{\infty} e^{inx}/n!= \sum_{n=0}^{\infty} (\cos nx + i\sin nx )/n!.$$
By equating the imaginary terms in both expressions we get 
 $$e^{\cos x}\sin (\sin x)=\sum_{n=0}^\infty \frac{\sin nx}{n!}$$
as required.