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

Σ_(r=0) ^∞ ((sin(rα))/(r!))

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

Commented by CAIMAN last updated on 27/Nov/21

alpha€C non?

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

Σ_(n=0) ^∞ ((sin(nα))/(n!))=Im(Σ_(n=0) ^∞ (e^(inα) /(n!))) we have Σ_(n=0) ^∞ (e^(inα) /(n!))=Σ_(n=0) ^∞ (((e^(iα) )^n )/(n!)) =e^e^(iα) =e^(cosα+isinα) =e^(cosα) (cos(sinα)+isin(sinα)) ⇒ Σ_(n=0) ^∞ ((sin(nα))/(n!))=e^(cosα) sin(sinα)

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

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