r-0-sin-r-r-

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

\underset{r = 0}{\sum^{\infty}} \frac{s i n (r α)}{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_{n = 0}^{\infty} \frac{\sin (n α)}{n!} = Im (\sum_{n = 0}^{\infty} \frac{e^{in α}}{n!}) we have

\sum_{n = 0}^{\infty} \frac{e^{in α}}{n!} = \sum_{n = 0}^{\infty} \frac{{(e^{i α})}^{n}}{n!} = e^{e^{i α}} = e^{\cos α + isin α}

= e^{\cos α} (\cos (\sin α) + isin (\sin α)) \Rightarrow

\sum_{n = 0}^{\infty} \frac{\sin (n α)}{n!} = e^{\cos α} \sin (\sin α)