prove-that-2-2-2-2cos-pi-2-n-

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 + \dots . + \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 + \dots}}}

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 \dots

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Sep/18

t h a n k y o u s i r \dots

Answered by behi83417@gmail.com last updated on 07/Sep/18

y = \sqrt{2 + \sqrt{2 + \sqrt{\dots \dots . . \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