calculate S_n =Σ_(k=0) ^(n−1) sin((π/(4n)) +((kπ)/(2n)))


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Question Number 51996 by maxmathsup by imad last updated on 01/Jan/19

$c a l c u l a t e S_{n} = \sum_{k = 0}^{n - 1} s i n (\frac{π}{4 n} + \frac{k π}{2 n})$
Commented by Abdo msup. last updated on 19/Jan/19

$w e h a v e S_{n} = \sum_{k = 0}^{n - 1} s i n (\frac{(2 k + 1) π}{4 n})$ $= I m (\sum_{k = 0}^{n - 1} e^{i (\frac{(2 k + 1) π}{4 n})}) b u t$ $A_{n} = \sum_{k = 0}^{n - 1} e^{i \frac{(2 k + 1) π}{4 n}} = e^{i \frac{π}{4 n}} \sum_{k = 0}^{n - 1} {(e^{\frac{i π}{2 n}})}^{k}$ $= e^{\frac{i π}{4 n}} \frac{1 - {(e^{\frac{i π}{2 n}})}^{n}}{1 - e^{\frac{i π}{2 n}}} = e^{\frac{i π}{4 n}} \frac{1 - i}{1 - c o s (\frac{π}{2 n}) - i s i n (\frac{π}{2 n})}$ $= e^{\frac{i π}{4 n}} \frac{\sqrt{2} e^{- \frac{i π}{4}}}{2 {s i n}^{2} (\frac{π}{4 n}) - 2 i s i n (\frac{π}{4 n}) c o s (\frac{π}{4 n})}$ $= \frac{\sqrt{2}}{2} e^{\frac{i π}{4 n}} \frac{e^{- \frac{i π}{4}}}{- i s i n (\frac{π}{4 n}) (e^{\frac{i π}{4 n}})}$ $= i \frac{\sqrt{2}}{2} e^{\frac{i π}{4 n}} \frac{e^{- \frac{i π}{4}} e^{- \frac{i π}{4 n}}}{s i n (\frac{π}{4 n})} = i \frac{\sqrt{2}}{2} \frac{\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}}{s i n (\frac{π}{4 n})}$ $= \frac{i}{2} \frac{1 - i}{s i n (\frac{π}{4 n})} = \frac{1}{2} \frac{i + 1}{s i n (\frac{π}{4 n})} \Rightarrow ★ S_{n} = \frac{1}{2 s i n (\frac{π}{4 n})} ★$
	Answered by tanmay.chaudhury50@gmail.com last updated on 02/Jan/19

	$T_{1} = s i n (\frac{π}{4 n}) = s i n (\frac{π}{4 n})$ $T_{2} = s i n (\frac{π}{4 n} + \frac{π}{2 n}) = s i n (\frac{3 π}{4 n})$ $T_{3} = s i n (\frac{π}{4 n} + \frac{2 π}{2 n}) = s i n (\frac{5 π}{4 n})$ $T_{4} = s i n (\frac{π}{4 n} + \frac{3 π}{2 n}) = s i n (\frac{7 π}{4 n})$ $. . .$ $. . . .$ $T_{n} = s i n (\frac{π}{4 n} + \frac{(n - 1) π}{2 n}) = s i n (\frac{π + 2 (n - 1) π}{4 n})$ $n o w$ $2 s i n (\frac{π}{4 n}) s i n (\frac{π}{4 n}) = c o s 0 - c o s (\frac{2 π}{4 n})$ $2 s i n (\frac{3 π}{4 n}) s i n (\frac{π}{4 n}) = c o s (\frac{2 π}{4 n}) - c o s (\frac{4 π}{4 n})$ $2 s i n (\frac{5 π}{4 n}) s i n (\frac{π}{4 n}) = c o s (\frac{4 π}{4 n}) - c o s (\frac{6 π}{4 n})$ $. . . .$ $. . . .$ $2 s i n (\frac{π + 2 (n - 1) π}{4 n}) s i n (\frac{π}{4 n}) = c o s (\frac{2 (n - 1) π}{4 n}) - c o s (\frac{2 n π}{4 n})$ $a d d t h e m$ $2 s i n (\frac{π}{4 n}) \times S_{n} = c o s 0 - c o s (\frac{2 n π}{4 n})$ $2 s i n (\frac{π}{4 n}) \times S_{n} = 1 - c o s (\frac{π}{2})$ $S_{n} = \frac{1}{2 s i n (\frac{π}{4 n})}$
	Commented by Abdo msup. last updated on 19/Jan/19

	$t h a n k s s i r T a n m a y .$
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