On pose H_n (α)=Π_(k=1) ^(n−1) sin((α/(2n))+((kπ)/n)) a) montrer que ∀α≠0, 2^(n−1) H_n (α)=((sin((α/2)))/(sin((α/(2n))))) b) Calculer lim_(α→0) H_n (α) c) De^� duire que ∀α≥2, sin((π/n))×sin(((2π)/n))×....×sin((((n−1)π)/n))=(π/2^(n−1) )..


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Question Number 147381 by puissant last updated on 20/Jul/21

$O n p o s e H_{n} (α) = \underset{k = 1}{\prod^{n - 1}} s i n (\frac{α}{2 n} + \frac{k π}{n})$ $a) m o n t r e r q u e \forall α \neq 0,$ $2^{n - 1} H_{n} (α) = \frac{s i n (\frac{α}{2})}{s i n (\frac{α}{2 n})}$ $b) C a l c u l e r {l i m}_{α \to 0} H_{n} (α)$ $c) D \overset{´}{e d u i r e} q u e \forall α ⩾ 2,$ $s i n (\frac{π}{n}) \times s i n (\frac{2 π}{n}) \times . . . . \times s i n (\frac{(n - 1) π}{n}) = \frac{π}{2^{n - 1}} . .$
	Answered by Olaf_Thorendsen last updated on 20/Jul/21

	$a)$ $P (z) = z^{n} - e^{i α}$ $P (z_{k}) = 0 \Leftrightarrow z_{k} = e^{i (\frac{α}{n} + \frac{2 k π}{n})}, k = 0, 1 . . ., n - 1$ $P (z) = \underset{k = 0}{\prod^{n - 1}} (z - z_{k})$ $P (z) = \underset{k = 0}{\prod^{n - 1}} (z - e^{i (\frac{α}{n} + \frac{2 k π}{n})})$ $P (1) = 1 - e^{i α} = \underset{k = 0}{\prod^{n - 1}} (1 - e^{i (\frac{α}{n} + \frac{2 k π}{n})})$ $1 - e^{i α} = (1 - e^{i \frac{α}{n}}) \underset{k = 1}{\prod^{n - 1}} (1 - e^{i (\frac{α}{n} + \frac{2 k π}{n})})$ $\underset{k = 1}{\prod^{n - 1}} (1 - e^{i (\frac{α}{n} + \frac{2 k π}{n})}) = \frac{1 - e^{i α}}{1 - e^{i \frac{α}{n}}}$ $C^{'} est fastidieux mais il suffit de$ $factoriser partout par l^{'} exponentielle$ $de la demi somme pour voir apparaitre$ $le resultat .$ $Par exemple 1 - e^{i α} est factorise en$ $e^{i \frac{α}{2}} (e^{- i \frac{α}{2}} - e^{i \frac{α}{2}}) . . . e t c . . .$
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