let p(x)=cos(2n arccos(x)) with x∈[−1,1] find the roots of p(x) and factorize p(x)


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Question Number 34020 by prof Abdo imad last updated on 29/Apr/18

$l e t p (x) = c o s (2 n a r c c o s (x)) w i t h x \in [- 1, 1]$ $f i n d t h e r o o t s o f p (x) a n d f a c t o r i z e p (x)$
Commented by math khazana by abdo last updated on 02/May/18
$let put arccosx=t ⇔ x=cost p(x)=0 ⇔ cos(2nt)=0 ⇔ 2nt =(π/2) +kπ k∈ Z ⇔ t_k = (π/(4n)) +((kπ)/(2n)) k ∈Z but card{t_k }<+∞ we can prove that k∈[0,2n−1]] so the roots of p(x)are x_k =cos(t_k ) =cos((π/(4n)) +((kπ)/(2n))) and p(x)=λ Π_(k=0) ^(2n−1) (x −cos((π/(4n)) +((kπ)/(2n)))) .$
$l e t p u t a r c c o s x = t \Leftrightarrow x = c o s t$ $p (x) = 0 \Leftrightarrow c o s (2 n t) = 0 \Leftrightarrow 2 n t = \frac{π}{2} + k π k \in Z$ $\Leftrightarrow t_{k} = \frac{π}{4 n} + \frac{k π}{2 n} k \in Z b u t c a r d {t_{k}} < + \infty w e c a n$ $p r o v e t h a t k \in [0, 2 n - 1]] s o t h e r o o t s o f p (x) a r e$ $x_{k} = c o s (t_{k}) = c o s (\frac{π}{4 n} + \frac{k π}{2 n}) a n d$ $p (x) = λ \prod_{k = 0}^{2 n - 1} (x - c o s (\frac{π}{4 n} + \frac{k π}{2 n})) .$
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