Use-De-Moivre-s-theorem-to-prove-that-r-0-cos-rx-1-2-and-find-a-value-or-expression-for-r-0-sin-rx-assume-that-this-two-series-were-convergent-

Question Number 136582 by physicstutes last updated on 23/Mar/21

Use {De}^{'} {Moivre}^{'} s theorem to prove that

\underset{r = 0}{\sum^{\infty}} \cos r x = \frac{1}{2} and find a value (or expression) for \underset{r = 0}{\sum^{\infty}} \sin r x

assume that this two series were convergent .

Answered by Ar Brandon last updated on 23/Mar/21

\underset{r = 0}{\sum^{\infty}} e^{irx} = \frac{1}{1 - e^{ix}} = \frac{e^{- \frac{x}{2} i}}{e^{- \frac{x}{2} i} - e^{\frac{x}{2} i}} = \frac{e^{- \frac{x}{2} i}}{- 2 isin \frac{x}{2}} = \frac{e^{- \frac{x}{2} i - \frac{3 π}{2} i}}{2 \sin \frac{x}{2}}

= \frac{\cos (\frac{x}{2} + \frac{3 π}{2}) - isin (\frac{x}{2} + \frac{3 π}{2})}{2 \sin \frac{x}{2}} = \frac{\sin \frac{x}{2} + icos \frac{x}{2}}{2 \sin \frac{x}{2}}

\underset{r = 0}{\sum^{\infty}} \cos (rx) = Re \underset{r = 0}{\sum^{\infty}} e^{irx} = \frac{1}{2}

\underset{r = 0}{\sum^{\infty}} \sin (rx) = Im \underset{r = 0}{\sum^{\infty}} e^{irx} = \frac{1}{2} \cot (\frac{x}{2})

Commented by physicstutes last updated on 23/Mar/21

perfect