prove that it exist one polynomial p/ p(cosx)=cos(nx) find the roots of p(x) .


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Question Number 30598 by abdo imad last updated on 23/Feb/18

$p r o v e t h a t i t e x i s t o n e p o l y n o m i a l p /$ $p (c o s x) = c o s (n x) f i n d t h e r o o t s o f p (x) .$
Commented by abdo imad last updated on 27/Feb/18

$w e h a v e b y m o i v r e f o r m u l a$ $c o s (n x) + i s i n (n x) = {(c o s x + i s i n x)}^{n}$ $= \sum_{k = 0}^{n} C_{n}^{k} {(i s i n x)}^{k} {(c o s x)}^{n - k}$ $= \sum_{p = 0}^{[\frac{n}{2}]} C_{n}^{2 p} {(i s i n x)}^{2 p} {(c o s x)}^{n - 2 p} + \sum_{p = 0}^{[\frac{n - 1}{2}]} = C_{n}^{2 p + 1} {(i s i n x)}^{2 p + 1} {(c o s x)}^{n - 2 p - 1}$ $c o s (n x) = R e (e^{i n x}) = \sum_{p = 0}^{[\frac{n}{2}]} {(- 1)}^{p} C_{n}^{2 p} {(1 - {c o s}^{2} x)}^{p} {(c o s x)}^{n - 2 p}$ $= p (c o s x) / p (x) = \sum_{p = 0}^{[\frac{n}{2}]} {(- 1)}^{p} C_{n}^{2 p} {(1 - x^{2})}^{p} x^{n - 2 p} .$ $2) X r o o t o f p (x) \Leftrightarrow p (X) = 0 l e t p u t X = c o s θ$ $p (X) = 0 \Leftrightarrow p (c o s θ) = 0 \Leftrightarrow c o s (n θ) = 0$ $\Leftrightarrow n θ = \frac{π}{2} + k π \Leftrightarrow θ = \frac{π}{2 n} + \frac{k π}{n} = \frac{(2 k + 1) π}{2 n} b u t w e c a n c h o w$ $t h a t d e g p = n \Rightarrow t h e r o i t s o f p (x) a r e$ $X_{k} = c o s (θ_{k}) = c o s ((2 k + 1) \frac{π}{2 n}) w i t h k \in [[0, n - 1]] .$
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