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Mathematica tells me that $$ \sum_{k=0}^\infty \sin(ak) \frac{x^k}{k!} = e^{x \cos (a)} \sin (x \sin (a)). $$ How would I go about evaluating such a series by hand? My first thought is to expand $\sin(ak)$, obtaining $$ \sum_{k=0}^\infty \sum_{n=0}^\infty \frac{(-1)^n (ak)^{2n+1} x^k}{k! (2 n+1)!} = \sum_{n=0}^\infty \frac{(-1)^n a^{2n+1}}{(2 n+1)!} \sum_{k=0}^\infty \frac{k^{2n+1} x^k}{k!} = e^x \sum_{n=0}^\infty \frac{(-1)^n a^{2n+1}}{(2 n+1)!} B_{2n+1}(x)$$ where $B_n(x)$ is a Bell polynomial. However, it is unclear to me how to proceed from here.

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    $\begingroup$ As soon as there is a power and a $\sin(ak)$ think De Moivre's formula. $\endgroup$ – Ali Caglayan Jul 16 '14 at 13:09
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$$ \sum_{k=0}^\infty \sin(ak) \frac{x^k}{k!} = \Im \left[\sum_{k=0}^\infty e^{iak} \frac{x^k}{k!}\right]\\ = \Im [e^{xe^{ia}}] = e^{x \cos (a)} \sin (x \sin (a)). $$

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  • $\begingroup$ Fantastic. I can't believe I didn't think of such a simple manipulation. $\endgroup$ – David Zhang Jul 16 '14 at 13:09

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