q-n-n-0-cos-x-2-n-i-study-the-variation-of-q-n-ii-show-that-cosx-sin2x-2sinx-x-0-pi-2-iii-deduce-that-q-n-1-2-n-1-sin2x-sin-x-2-n-iv-lim-n-q-n-v-s

Question Number 147528 by alcohol last updated on 21/Jul/21

q_{n} = \underset{n = 0}{\prod^{\infty}} c o s (\frac{x}{2^{n}})

i) s t u d y t h e v a r i a t i o n o f q_{n}

i i)

s h o w t h a t c o s x = \frac{s i n 2 x}{2 s i n x}, \forall x \in [0, \frac{π}{2}]

i i i)

d e d u c e t h a t q_{n} = \frac{1}{2^{n + 1}} \times \frac{s i n 2 x}{s i n (\frac{x}{2^{n}})}

i v) \lim_{n \to \infty} q_{n} = ?

v)

s o l v e c o s (\frac{x}{2}) ⩾ - \frac{1}{2}

Commented by Olaf_Thorendsen last updated on 21/Jul/21

u s e \sin 2 θ_{n} = 2 \sin θ_{n} \cos θ_{n}

\Rightarrow \cos θ_{n} = \frac{\sin 2 θ_{n}}{2 \sin θ_{n}} with θ_{n} = \frac{x}{2^{n}}

\cos \frac{x}{2^{n}} = \frac{\sin \frac{x}{2^{n - 1}}}{2 \sin \frac{x}{2^{n}}}

\dots t e l e s c o p i c p r o d u c t

Commented by alcohol last updated on 21/Jul/21

t h a n k s

b u t i d o n t u n d e r s t a n d

Commented by puissant last updated on 21/Jul/21

U_{n} = \underset{k = 0}{\prod^{n}} \cos (\frac{x}{2^{k}}) \dots