prove that (√(2+(√(2+....+(√2))))) =2cos((π/2^n ))


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Question Number 43000 by abdo.msup.com last updated on 06/Sep/18

$p r o v e t h a t \sqrt{2 + \sqrt{2 + . . . . + \sqrt{2}}} = 2 c o s (\frac{π}{2^{n}})$
	Answered by tanmay.chaudhury50@gmail.com last updated on 06/Sep/18

	$c o s 2 α = 2 {c o s}^{2} α - 1$ ${c o s}^{2} α = \frac{1 + c o s 2 α}{2}$ $2 c o s α = 2 \sqrt{\frac{1 + c o s 2 α}{2}}$ $2 c o s α = \sqrt{2 (1 + c o s 2 α)}$ $2 c o s \frac{Π}{2^{2}} = \sqrt{2 (1 + c o s \frac{Π}{2})} = \sqrt{2}$ $2 c o s \frac{Π}{2^{3}} = \sqrt{2 (1 + c o s \frac{Π}{2^{2}})} = \sqrt{2 (1 + \frac{1}{\sqrt{2}})} = \sqrt{2 + \sqrt{2}}$ $2 c o s \frac{Π}{2^{4}} = \sqrt{2 (1 + c o s \frac{Π}{2^{3}})} = \sqrt{2 (1 + \frac{\sqrt{2 + \sqrt{2}}}{2})} = \sqrt{2 + \sqrt{2 + \sqrt{2}}}$ $t h u s 2 c o s \frac{Π}{2^{n}} = \sqrt{2 + \sqrt{2 + \sqrt{2 + . . .}}}$ $h e n c e p r o v e d$ $p l s c h e c k$
	Commented by maxmathsup by imad last updated on 06/Sep/18

	$y e s c o r r e c t i c a n s a y t h a t y o u h a v e u s i n g a r e c u r r e n c e p r o o f . . .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 06/Sep/18

	$t h a n k y o u s i r . . .$
	Answered by behi83417@gmail.com last updated on 07/Sep/18

	$y = \sqrt{2 + \sqrt{2 + \sqrt{. . . . . . . . \sqrt{2}}}}$ $y^{2} = 2 + y \Rightarrow y^{2} - y - 2 = 0$ $y = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 + 3}{2} = 2$ $\lim_{n \to \infty} (2 c o s \frac{π}{2^{n}}) = 2$
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