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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.

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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.

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  • $\begingroup$ Perfect, thanks a lot. $\endgroup$ – Jozef Jan 9 '12 at 15:26
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Try thinking about the imaginary part of $e^{e^{i x}}$ in two different ways.

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    $\begingroup$ Can you please extend your answer? I can't see it nor understand it. Thanks a lot! $\endgroup$ – Jozef Jan 9 '12 at 8:01

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