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Question Number 108870 by 150505R last updated on 19/Aug/20

Commented by Dwaipayan Shikari last updated on 19/Aug/20

$c o t \frac{π}{2 n}$
	Answered by Ar Brandon last updated on 19/Aug/20
	$cos((πk)/n)+isin((πk)/n)=e^(i((πk)/n)) ⇒Σ_(k=1) ^(n−1) sin((πk)/n)=ImΣ_(k=1) ^(n−1) e^(i((πk)/n)) =Im{((e^(i(π/n)) (1−e^(iπ) ))/(1−e^((iπ)/n) ))}$
	$\cos \frac{π k}{n} + isin \frac{π k}{n} = e^{i \frac{π k}{n}}$ $\Rightarrow \underset{k = 1}{\sum^{n - 1}} \sin \frac{π k}{n} = Im \underset{k = 1}{\sum^{n - 1}} e^{i \frac{π k}{n}}$ $= Im {\frac{e^{i \frac{π}{n}} (1 - e^{i π})}{1 - e^{\frac{i π}{n}}}}$
	Answered by Dwaipayan Shikari last updated on 19/Aug/20

	$\underset{k = 0}{\sum^{n}} s i n k θ (θ = (\frac{π}{n}))$ $s i n 0 + s i n θ + s i n 2 θ + . . . . + n$ $\frac{1}{2 s i n \frac{θ}{2}} (c o s \frac{θ}{2} - c o s \frac{θ}{2} + c o s \frac{θ}{2} - c o s \frac{3 θ}{2} + c o s \frac{3 θ}{2} - c o s \frac{5 θ}{2} + . . c o s \frac{2 n - 1}{2} θ - c o s \frac{2 n + 1}{2} θ)$ $\frac{1}{2 s i n \frac{θ}{2}} (c o s \frac{θ}{2} - c o s \frac{2 n + 1}{2} θ)$ $\frac{1}{s i n \frac{θ}{2}} (s i n \frac{n + 1}{2} θ s i n \frac{n}{2} θ) = \frac{1}{s i n \frac{π}{2 n}} s i n (\frac{π}{2 n} + \frac{π}{2}) . s i n \frac{π}{2} = c o t \frac{π}{2 n} (θ = \frac{π}{n})$
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