find-the-value-of-k-1-n-1-sin-kpi-2n-

Question Number 27153 by abdo imad last updated on 02/Jan/18

f i n d t h e v a l u e o f \prod_{k = 1}^{n - 1} s i n (\frac{k π}{2 n}) .

Commented by abdo imad last updated on 03/Jan/18

l e t i n t r o d u c e t h e p o l y n o m i a l p (x) = x^{2 n} - 1 t h e r o o t s o f p (x)

a r e t h e c o m p l e x z_{k} = e^{i \frac{2 k}{n}} a n d k \in [[0, 2 n - 1]] a n d

p (x) = λ \prod_{k = 0}^{n - 1} (x - z_{k}) i t s c l e a r t h a t λ = 1 a n d

p (x) = \prod_{k = 0}^{n - 1} (x - z_{k}) w e h a v e z_{0} = 1, z_{1} = e^{i \frac{π}{n}}, z_{2} = e^{i \frac{2 π}{n}}

z_{n - 1} = e^{i \frac{(n - 1) π}{n}}, z_{n} = - 1, z_{n + 1} = e^{i \frac{(n + 1) π}{n}}, z_{2 n - 1} = e^{i \frac{(2 n - 1) π}{2 n}}

w e s e e t h a t z_{2 n - 1} = z_{1}^{-}, z_{2 n - 2} = z_{2}^{-}, z_{n + 1} = z_{n - 1}^{-}

\Rightarrow p (x) = (x^{2} - 1) \prod_{k = 1}^{n - 1} (x - z_{k}) (x - z_{k}^{-})

= (x^{2} - 1) \prod_{k = 1}^{n - 1} (x^{2} - 2 c o s (\frac{k π}{n}) x + 1) a n d f o r x^{2} \neq 1

\frac{p (x)}{x^{2} - 1} = \prod_{k = 1}^{n - 1} (x^{2} - 2 c o s (\frac{k π}{n}) x + 1) a n d b y u s i n g h o s p i t a l t h e o e m

{l i m}_{x - > 1} \prod_{k = 1}^{n - 1} (x^{2} - 2 c o s (\frac{k π}{n}) x + 1) = {l i m}_{x - > 1} \frac{p^{^{'}} (x)}{2 x}

\prod_{k = 1}^{n - 1} 2 (1 - c o s (\frac{k π}{n})) = {l i m}_{x - > 1} \frac{2 {n x}^{2 n - 1}}{2 x} = n

\Rightarrow 4^{n - 1} \prod_{k = 1}^{k = n - 1} {s i n}^{2} (\frac{k π}{2 n}) = n

\Rightarrow \prod_{k = 1}^{n - 1} {s i n}^{2} (\frac{k π}{2 n}) = \frac{n}{4^{n - 1}}

\Rightarrow \prod_{k = 1}^{n - 1} s i n (\frac{k π}{2 n}) = \frac{\sqrt{n}}{2^{n - 1}} (n ⩾ 2)

Commented by Tinkutara last updated on 03/Jan/18

I have already asked this a long time back. See this at Q 19292.