Sum the series to n terms sin θ−sin 2θ+sin 3θ−........


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 142105 by PRITHWISH SEN 2 last updated on 26/May/21

$Sum the series to n terms$ $\sin θ - \sin 2 θ + \sin 3 θ - . . . . . . . .$
	Answered by Dwaipayan Shikari last updated on 26/May/21

	$\underset{n = 1}{\sum^{\infty}} {(- 1)}^{n} \frac{s i n (n θ)}{n} = \frac{1}{2 i} l o g (\frac{1 + e^{i θ}}{1 + e^{- i θ}}) = \frac{θ}{2}$ $\underset{n = 1}{\sum^{\infty}} {(- 1)}^{n} s i n (n θ) = \frac{1}{2}$ $s i n θ - s i n 2 θ + s i n 3 θ - . . . = - \frac{1}{2}$
	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
	$S_n =Σ_(k=1) ^n (−1)^(k+1) sin(kθ) ⇒S_n =−Σ_(k=1) ^n (−1)^k sin(kθ) =−Im(Σ_(k=1) ^n (−e^(iθ) )^k ) we have Σ_(k=1) ^n (−e^(iθ) )^k =Σ_(k0) ^n (−e^(iθ) )^k −1 =((1−(−e^(iθ) )^(n+1) )/(1+e^(iθ) ))−1 =((1−(−1)^(n+1) e^(i(n+1)θ) )/(1+e^(iθ) ))−1 =((1−e^(iπ(n+1)+i(n+1)θ) )/(1+e^(iθ) ))−1 =((1−e^(i(n+1)(θ+π)) )/(1+e^(iθ) ))−1 =((1−cos(n+1)(θ+π)−isin(n+1)(θ +π))/(1+cosθ+isinθ))−1 =((2sin^2 ((((n+1)(θ+π))/2))−2isin((((n+1)(θ+π))/2))cos((((n+1)(θ+π))/2)))/(2cos^2 ((θ/2))+2isin((θ/2))cos((θ/2))))−1 =((−isin((((n+1)(θ+π))/2))e^(i((((n+1)(θ +π))/2))) )/(cos((θ/2))e^(i(θ/2)) ))−1 =−i((sin((((n+1)(θ+π))/2)))/(cos((θ/2))))e^(i((((n+1)(θ+π))/2)−(θ/2))) −1 =−i((sin((((n+1)(θ+π))/2)))/(cos((θ/2)))){cos((((n+1)(θ+π)−θ)/2))+isin((((n+1)(θ+π)/(2 )))}−1 ⇒−Im(Σ...)=((sin((((n+1)(θ+π))/2)))/(cos((θ/2))))×cos((((n+1)(θ+π)−θ)/2)) =S_n$
	$S_{n} = \sum_{k = 1}^{n} {(- 1)}^{k + 1} \sin (k θ) \Rightarrow S_{n} = - \sum_{k = 1}^{n} {(- 1)}^{k} \sin (k θ)$ $= - Im (\sum_{k = 1}^{n} {(- e^{i θ})}^{k}) we have$ $\sum_{k = 1}^{n} {(- e^{i θ})}^{k} = \sum_{k 0}^{n} {(- e^{i θ})}^{k} - 1 = \frac{1 - {(- e^{i θ})}^{n + 1}}{1 + e^{i θ}} - 1$ $= \frac{1 - {(- 1)}^{n + 1} e^{i (n + 1) θ}}{1 + e^{i θ}} - 1$ $= \frac{1 - e^{i π (n + 1) + i (n + 1) θ}}{1 + e^{i θ}} - 1 = \frac{1 - e^{i (n + 1) (θ + π)}}{1 + e^{i θ}} - 1$ $= \frac{1 - \cos (n + 1) (θ + π) - isin (n + 1) (θ + π)}{1 + \cos θ + isin θ} - 1$ $= \frac{{2 \sin}^{2} (\frac{(n + 1) (θ + π)}{2}) - 2 isin (\frac{(n + 1) (θ + π)}{2}) \cos (\frac{(n + 1) (θ + π)}{2})}{{2 \cos}^{2} (\frac{θ}{2}) + 2 isin (\frac{θ}{2}) \cos (\frac{θ}{2})} - 1$ $= \frac{- isin (\frac{(n + 1) (θ + π)}{2}) e^{i (\frac{(n + 1) (θ + π)}{2})}}{\cos (\frac{θ}{2}) e^{i \frac{θ}{2}}} - 1$ $= - i \frac{\sin (\frac{(n + 1) (θ + π)}{2})}{\cos (\frac{θ}{2})} e^{i (\frac{(n + 1) (θ + π)}{2} - \frac{θ}{2})} - 1$ $= - i \frac{\sin (\frac{(n + 1) (θ + π)}{2})}{\cos (\frac{θ}{2})} {\cos (\frac{(n + 1) (θ + π) - θ}{2}) + isin (\frac{(n + 1) (θ + π}{2})} - 1$ $\Rightarrow - Im (Σ . . .) = \frac{\sin (\frac{(n + 1) (θ + π)}{2})}{\cos (\frac{θ}{2})} \times \cos (\frac{(n + 1) (θ + π) - θ}{2})$ $= S_{n}$
	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

	$you arewelcome sir .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum