prove that ((2 cos 2^n θ + 1)/(2 cos θ + 1)) = (2 cos θ − 1)(2 cos 2θ − 1)(2 cos 2^2 θ− 1) ...(2 cos 2^(n − 1) θ − 1)


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Question Number 38557 by kunal1234523 last updated on 27/Jun/18

$p r o v e t h a t$ $\frac{2 \cos 2^{n} θ + 1}{2 \cos θ + 1} = (2 \cos θ - 1) (2 \cos 2 θ - 1) (2 \cos 2^{2} θ - 1)$ $. . . (2 \cos 2^{n - 1} θ - 1)$
	Answered by kunal1234523 last updated on 27/Jun/18

	$2 c o s 2^{x} a + 1$ $= 2 c o s 2 (2^{x - 1}) a + 1$ $= 2 (2 {c o s}^{2} 2^{x - 1} a - 1) + 1$ $= 4 {c o s}^{2} 2^{x - 1} a - 2 + 1$ $= {(2 c o s 2^{x - 1} a)}^{2} - {(1)}^{2}$ $= (2 c o s 2^{x - 1} a + 1) (2 c o s 2^{x - 1} a - 1) . . . . . . . . (1)$ $L . H . S$ $\frac{2 \cos 2^{n} θ + 1}{2 \cos θ + 1}$ $f r o m (1), t a k i n g x = n$ $= \frac{(2 c o s 2^{n - 1} a + 1) (2 c o s 2^{n - 1} a - 1)}{2 c o s a + 1}$ $n o w f r o m (1), t a k i n g x = n - 1$ $= \frac{(2 c o s 2^{n - 2} a + 1) (2 c o s 2^{n - 2} a - 1) (2 c o s 2^{n - 1} a - 1)}{2 c o s a + 1}$ $a f t e r s o m e t i m e$ $= \frac{(2 c o s 2^{n - n} a + 1) (2 c o s 2^{n - n} a - 1) . . . . . . . . . . . . (2 c o s 2^{n - 3} a - 1) (2 c o s 2^{n - 2} a - 1) (2 c o s 2^{n - 1} a - 1)}{2 c o s a + 1}$ $= \frac{(2 c o s a + 1) (2 c o s a - 1) (2 c o s 2 a - 1) (2 c o s 2^{2} a - 1) . . . . . . . . . . . . (2 c o s 2^{n - 3} a - 1) (2 c o s 2^{n - 2} a - 1) (2 c o s 2^{n - 1} a - 1)}{(2 c o s a + 1)}$ $= (2 c o s a - 1) (2 c o s 2 a - 1) (2 c o s 2^{2} a - 1) . . . . . . . . . . . . (2 c o s 2^{n - 3} a - 1) (2 c o s 2^{n - 2} a - 1) (2 c o s 2^{n - 1} a - 1)$ $p r o v e d$
	Commented by kunal1234523 last updated on 05/Jul/18

	$i s t h i s r i g h t s i r$
	Commented by math khazana by abdo last updated on 06/Jul/18

	$i t s s e e m s t h a t y o u a n s w e r i s c o r r e c t i f y o u h a v e$ $u s e d r e c u r r e n c e r u l e .$
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