Sum-the-series-to-n-terms-sin-sin-2-sin-3-

Question Number 142105 by PRITHWISH SEN 2 last updated on 26/May/21

Sum the series to n terms

\sin θ - \sin 2 θ + \sin 3 θ - \dots \dots . .

Answered by Dwaipayan Shikari last updated on 26/May/21

\underset{n = 1}{\sum^{\infty}} {(- 1)}^{n} \frac{s i n (n θ)}{n} = \frac{1}{2 i} l o g (\frac{1 + e^{i θ}}{1 + e^{- i θ}}) = \frac{θ}{2}

\underset{n = 1}{\sum^{\infty}} {(- 1)}^{n} s i n (n θ) = \frac{1}{2}

s i n θ - s i n 2 θ + s i n 3 θ - \dots = - \frac{1}{2}

Commented by PRITHWISH SEN 2 last updated on 26/May/21

thank you sir

Answered by mathmax by abdo last updated on 26/May/21

S_{n} = \sum_{k = 1}^{n} {(- 1)}^{k + 1} \sin (k θ) \Rightarrow S_{n} = - \sum_{k = 1}^{n} {(- 1)}^{k} \sin (k θ)

= - Im (\sum_{k = 1}^{n} {(- e^{i θ})}^{k}) we have

\sum_{k = 1}^{n} {(- e^{i θ})}^{k} = \sum_{k 0}^{n} {(- e^{i θ})}^{k} - 1 = \frac{1 - {(- e^{i θ})}^{n + 1}}{1 + e^{i θ}} - 1

= \frac{1 - {(- 1)}^{n + 1} e^{i (n + 1) θ}}{1 + e^{i θ}} - 1

= \frac{1 - e^{i π (n + 1) + i (n + 1) θ}}{1 + e^{i θ}} - 1 = \frac{1 - e^{i (n + 1) (θ + π)}}{1 + e^{i θ}} - 1

= \frac{1 - \cos (n + 1) (θ + π) - isin (n + 1) (θ + π)}{1 + \cos θ + isin θ} - 1

= \frac{{2 \sin}^{2} (\frac{(n + 1) (θ + π)}{2}) - 2 isin (\frac{(n + 1) (θ + π)}{2}) \cos (\frac{(n + 1) (θ + π)}{2})}{{2 \cos}^{2} (\frac{θ}{2}) + 2 isin (\frac{θ}{2}) \cos (\frac{θ}{2})} - 1

= \frac{- isin (\frac{(n + 1) (θ + π)}{2}) e^{i (\frac{(n + 1) (θ + π)}{2})}}{\cos (\frac{θ}{2}) e^{i \frac{θ}{2}}} - 1

= - i \frac{\sin (\frac{(n + 1) (θ + π)}{2})}{\cos (\frac{θ}{2})} e^{i (\frac{(n + 1) (θ + π)}{2} - \frac{θ}{2})} - 1

= - i \frac{\sin (\frac{(n + 1) (θ + π)}{2})}{\cos (\frac{θ}{2})} {\cos (\frac{(n + 1) (θ + π) - θ}{2}) + isin (\frac{(n + 1) (θ + π}{2})} - 1

\Rightarrow - Im (Σ \dots) = \frac{\sin (\frac{(n + 1) (θ + π)}{2})}{\cos (\frac{θ}{2})} \times \cos (\frac{(n + 1) (θ + π) - θ}{2})

= S_{n}

Commented by PRITHWISH SEN 2 last updated on 27/May/21

Thank you sir

Commented by mathmax by abdo last updated on 28/May/21

you arewelcome sir .