Prove that, for any real number x and odd positive integer n, cos^n x=(1/2^(n+1) )Σ_(k=0) ^((n−1)/2) C


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Question Number 170800 by thfchristopher last updated on 31/May/22

$Prove that, for any real number x and odd positive integer n,$ $\cos^{n} x = \frac{1}{2^{n + 1}} \underset{k = 0}{\sum^{(n - 1) / 2}} C_{k}^{n} \cos (n - 2 k) x$
	Answered by aleks041103 last updated on 04/Jun/22

	$c o s x = \frac{e^{i x} + e^{- i x}}{2}$ ${c o s}^{n} x = \frac{1}{2^{n}} {(e^{i x} + e^{- i x})}^{n} =$ $= \frac{1}{2^{n}} (\underset{k = 0}{\sum^{n}} C_{k}^{n} e^{i k x} e^{- i (n - k) x}) =$ $= \frac{1}{2^{n}} (\underset{k = 0}{\sum^{n}} C_{k}^{n} e^{i x (2 k - n)})$ $\underset{k = 0}{\sum^{n}} C_{k}^{n} e^{i x (2 k - n)} =$ $= \underset{k = 0}{\sum^{(n - 1) / 2}} C_{k}^{n} e^{i x (2 k - n)} + \underset{k = (n + 1) / 2}{\sum^{n}} C_{k}^{n} e^{i x (2 k - n)}$ $\underset{k = (n + 1) / 2}{\sum^{n}} C_{k}^{n} e^{i x (2 k - n)}$ $l e t n - 2 j = 2 k - n$ $\Rightarrow 2 j = 2 n - 2 k \Rightarrow j = n - k, k = n - j$ $j_{1} = n - \frac{n + 1}{2} = \frac{n - 1}{2}$ $j_{2} = n - n = 0$ $\Rightarrow \underset{k = (n + 1) / 2}{\sum^{n}} C_{k}^{n} e^{i x (2 k - n)} = \underset{j = 0}{\sum^{(n - 1) / 2}} C_{n - j}^{n} e^{- i x (2 j - n)}$ $b u t C_{n - j}^{n} = C_{j}^{n}$ $c h a n g i n g j t o k$ $\Rightarrow \underset{k = (n + 1) / 2}{\sum^{n}} C_{k}^{n} e^{i x (2 k - n)} = \underset{k = 0}{\sum^{(n - 1) / 2}} C_{k}^{n} e^{- i x (2 k - n)}$ $\Rightarrow \underset{k = 0}{\sum^{n}} C_{k}^{n} e^{i x (2 k - n)} =$ $= \underset{k = 0}{\sum^{(n - 1) / 2}} C_{k}^{n} e^{i x (2 k - n)} + \underset{k = 0}{\sum^{(n - 1) / 2}} C_{k}^{n} e^{- i x (2 k - n)} =$ $= \underset{k = 0}{\sum^{(n - 1) / 2}} C_{k}^{n} (e^{i x (2 k - n)} + e^{- i x (2 k - n)}) =$ $= 2 \underset{k = 0}{\sum^{(n - 1) / 2}} C_{k}^{n} c o s ((2 k - n) x)$ $\Rightarrow {c o s}^{n} x = \frac{1}{2^{n - 1}} \underset{k = 0}{\sum^{(n - 1) / 2}} C_{k}^{n} c o s ((2 k - n) x)$ $c o s i s e v e n f u n c t i o n$ ${c o s}^{n} x = \frac{1}{2^{n - 1}} \underset{k = 0}{\sum^{(n - 1) / 2}} C_{k}^{n} c o s ((n - 2 k) x)$
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