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Question Number 160263 by alcohol last updated on 26/Nov/21

$$\underset{r=0}{\sum ^{\mathrm{\infty }}}\frac{sin\left(r\alpha \right)}{r!}$$

Commented by CAIMAN last updated on 27/Nov/21

alpha€C non?

Answered by mathmax by abdo last updated on 27/Nov/21

$$\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{\sin \left(\mathrm{n}\alpha \right)}{\mathrm{n}!}=\mathrm{Im}\left(\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{in}\alpha }}{\mathrm{n}!}\right)\mathrm{we}\mathrm{have}$$$$\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{in}\alpha }}{\mathrm{n}!}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\left({\mathrm{e}}^{\mathrm{i}\alpha }\right)}^{\mathrm{n}}}{\mathrm{n}!}={\mathrm{e}}^{{\mathrm{e}}^{\mathrm{i}\alpha }}={\mathrm{e}}^{\cos \alpha +\mathrm{isin}\alpha }$$$$={\mathrm{e}}^{\cos \alpha }(\cos \left(\sin \alpha \right)+\mathrm{isin}\left(\sin \alpha \right))\Rightarrow$$$$\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{\sin \left(\mathrm{n}\alpha \right)}{\mathrm{n}!}={\mathrm{e}}^{\cos \alpha }\sin \left(\sin \alpha \right)$$$$$$

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