montrer-que-2cos-pi-2-n-2-2-2-

Question Number 153009 by SANOGO last updated on 04/Sep/21

m o n t r e r q u e :

2 c o s \frac{π}{2^{n}} = \sqrt{2 + \sqrt{2 + \dots + \sqrt{2}}}

Commented by mr W last updated on 04/Sep/21

p l e a s e m a k e y o u r p o s t s b e t t e r r e a d a b l e!

t r y t o s e t t h e d e f a u t f o n t s i z e 20 .

t r y t o a v o i d u s i n g e x t r e m l y l a r g e o r

e x t r e m l y s m a l l f o n t s!

Commented by SANOGO last updated on 04/Sep/21

o k t h a n k y o u i u n d e r s t o o d

Answered by puissant last updated on 04/Sep/21

m o n t r o n s a l o r s q u e \forall n \in N, 2 c o s \frac{π}{2^{n}} = \sqrt{2 + \sqrt{2 + \dots . + \sqrt{2}}}

\Rightarrow c o s \frac{π}{2^{n}} = \frac{1}{2} \sqrt{2 + \sqrt{2 + \sqrt{2 + . . \sqrt{2}}}}

\to p o u r n = 1 l a p r o p r i e t \overset{´}{e} e s t v r a i e . .

\to s o i t n \in N, s u p p o s o n s l a p r o p r i e t \overset{´}{e} v r a i e

a u r a n g n e t p r o u v o n s a u r a n g n + 1 .

R e m a r q u o n s a u p r \overset{´}{e a l a b l e} q u e \forall x \in [0; \frac{π}{2}]

c o m m e {c o s}^{2} x = \frac{c o s 2 x + 1}{2} \Rightarrow c o s (\frac{x}{2}) = \sqrt{\frac{c o s x + 1}{2}}

e t c o s (\frac{π}{2^{n + 1}}) = c o s (\frac{1}{2} (\frac{π}{2^{n}}))

= \sqrt{\frac{1}{2} (c o s (\frac{π}{2^{n}}) + 1)}

= \sqrt{\frac{1}{2} (\frac{1}{2} \sqrt{2 + \sqrt{2 + \dots + \sqrt{2}}} + 1)} o n a a i n s i

n r a d i c a u x d^{'} a p r e s l^{'} h y p o t h e s e d e r e c c u r e n c e .

= \frac{1}{2} \sqrt{2 + \sqrt{2 + \dots + \sqrt{2}}} (n + 1 r a d i c a u x)

o n p r o u v e a i n s i l a p r o p r i e t \overset{´}{e} p a r r e c c u r e n c e . .

Commented by SANOGO last updated on 04/Sep/21

m e r c i b i e n q u e D i e u t e b e n i s s e

Answered by som(math1967) last updated on 04/Sep/21

c o s \frac{π}{4} = \frac{1}{\sqrt{2}}

2 c o s \frac{π}{4} = \sqrt{2} ∴ 2 c o s \frac{π}{2^{2}} = \sqrt{2}

c o s \frac{π}{8} = \sqrt{\frac{1 + c o s \frac{π}{4}}{2}} [∵ c o s \frac{θ}{2} = \sqrt{\frac{1 + c o s θ}{2}}]

c o s \frac{π}{8} = \sqrt{\frac{\sqrt{2} + 1}{2 \sqrt{2}}} = \sqrt{\frac{2 + \sqrt{2}}{4}} =

2 c o s \frac{π}{2^{3}} = \sqrt{2 + \sqrt{2}}

s a m e w a y 2 c o s \frac{π}{2^{4}} = \sqrt{2 + \sqrt{2 + \sqrt{2}}}

2 c o s \frac{π}{2^{5}} = \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2}}}}

\dots \dots . 2 c o s \frac{π}{2^{n}} = \sqrt{2 + \sqrt{2 + \sqrt{2 + \dots \sqrt{2}}}}

Commented by SANOGO last updated on 04/Sep/21

m e r c i b e a u c o u l e d u r

Answered by Jonathanwaweh last updated on 05/Sep/21

m o n t r e r q u :

2 c o s \frac{π}{2^{n}} = \sqrt{2 + \sqrt{2 + \dots + \sqrt{2}}}

p o s o n s U_{n} = \sqrt{2 + \sqrt{2 + \dots + \sqrt{2}}} a l o r s U_{n + 1} = \sqrt{U_{n} + 2}

s o i t p (n) : \forall n ϵ N n ⩾ 2 2 c o s \frac{Π}{2^{n}} = U_{n}

1 - i n i t i a l i s a t i o n (f a c i l e)

2 - h e r e d i t e

p o u r n ⩾ 2 {f i x e}^{^{'}} s u p p o s o n s p (n) v r a i e e t m o n t r o n s q u e p (n + 1) l^{'} e s t a u s s i

i e 2 c o s \frac{Π}{2^{n + 1}} = U_{n + 1} . o n a c o s (\frac{Π}{2^{n}}) = c o s (\frac{Π}{2^{n + 1}} \times 2)

= 2 {c o s}^{2} (\frac{Π}{2^{n + 1}}) - 1

d o n c 2 c o s (\frac{Π}{2^{n + 1}}) = 4 {c o s}^{2} (\frac{Π}{2^{n + 1}}) - 2

i e \sqrt{2 c o s (\frac{Π}{2^{n}}) + 2} = 2 c o s (\frac{Π}{2^{n + 1}})

d^{^{'}} o u o n a U_{n + 1} = 2 c o s (\frac{Π}{2^{n + 1}})

c o n c l u s i o n : \dots

Commented by SANOGO last updated on 05/Sep/21

m e r c i b i e n m o n c h e f

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