sin A+sin 2A+sin 3A+...+sin nA =??


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Question Number 114627 by john santu last updated on 20/Sep/20

$\sin A + \sin 2 A + \sin 3 A + . . . + \sin n A = ? ?$
	Answered by mathmax by abdo last updated on 20/Sep/20
	$let S_n (x) =sinx +sin(2x)+...+sin(nx)=Σ_(k=0) ^n sin(kx) =Im(Σ_(k=0) ^n e^(ikx) ) we have Σ_(k=0) ^n e^(ikx) =Σ_(k=0) ^n (e^(ix) )^k =((1−(e^(ix) )^(n+1) )/(1−e^(ix) )) =((1−cos(n+1)x−isin(n+1)x)/(1−cosx −isinx)) =((2sin^2 (n+1)(x/2)−2i sin((n+1)(x/2))cos(n+1)(x/2))/(2sin^2 ((x/2))−2isin((x/2))cos((x/2)))) =((−isin(n+1)(x/2){ e^(i(n+1)(x/2)) })/(−isin((x/2))e^((ix)/2) )) =((sin((n+1)(x/2)))/(sin((x/2))))×e^(in(x/2)) =((sin((((n+1)x)/2)))/(sin((x/2))))×{cos(((nx)/2))+isin(((nx)/2))} ⇒ S_n (x) =((sin(((nx)/2))sin((((n+1)x)/2)))/(sin((x/2)))) (x≠2kπ )$
	$let S_{n} (x) = sinx + \sin (2 x) + . . . + \sin (nx) = \sum_{k = 0}^{n} \sin (kx)$ $= Im (\sum_{k = 0}^{n} e^{ikx}) we have \sum_{k = 0}^{n} e^{ikx} = \sum_{k = 0}^{n} {(e^{ix})}^{k} = \frac{1 - {(e^{ix})}^{n + 1}}{1 - e^{ix}}$ $= \frac{1 - \cos (n + 1) x - isin (n + 1) x}{1 - cosx - isinx}$ $= \frac{{2 \sin}^{2} (n + 1) \frac{x}{2} - 2 i \sin ((n + 1) \frac{x}{2}) \cos (n + 1) \frac{x}{2}}{{2 \sin}^{2} (\frac{x}{2}) - 2 isin (\frac{x}{2}) \cos (\frac{x}{2})}$ $= \frac{- isin (n + 1) \frac{x}{2} {e^{i (n + 1) \frac{x}{2}}}}{- isin (\frac{x}{2}) e^{\frac{ix}{2}}} = \frac{\sin ((n + 1) \frac{x}{2})}{\sin (\frac{x}{2})} \times e^{in \frac{x}{2}}$ $= \frac{\sin (\frac{(n + 1) x}{2})}{\sin (\frac{x}{2})} \times {\cos (\frac{nx}{2}) + isin (\frac{nx}{2})} \Rightarrow$ $S_{n} (x) = \frac{\sin (\frac{nx}{2}) \sin (\frac{(n + 1) x}{2})}{\sin (\frac{x}{2})} (x \neq 2 k π)$
	Answered by Dwaipayan Shikari last updated on 20/Sep/20

	$s i n A + s i n 2 A + . . . . .$ $\frac{1}{2 s i n \frac{A}{2}} (c o s \frac{A}{2} - c o s \frac{3 A}{2} + c o s \frac{3 A}{2} - c o s \frac{5 A}{2} + c o s \frac{5 A}{2} - c o s \frac{7 A}{2} + . . . . c o s \frac{2 n - 1}{2} A - c o s \frac{2 n + 1}{2} A)$ $\frac{1}{2 s i n \frac{A}{2}} (2 s i n \frac{n + 1}{2} A . s i n \frac{n}{2} A) = \frac{s i n \frac{n + 1}{2} A . s i n \frac{n}{2} A}{s i n \frac{A}{2}}$
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