Prove-that-x-pi-4-n-0-cos-x-2-n-cos-pi-2x-2-n-1-2-4cos2x-pi-pi-4x-

Question Number 83209 by ~blr237~ last updated on 28/Feb/20

P r o v e t h a t \forall x \neq \frac{π}{4}, \underset{n = 0}{\prod^{\infty}} (\frac{c o s (\frac{x}{2^{n}}) + c o s (\frac{π - 2 x}{2^{n + 1}})}{2}) = \frac{4 c o s 2 x}{π (π - 4 x)}

Answered by mind is power last updated on 28/Feb/20

c o s (a) + c o s (b) = 2 c o s (\frac{a + b}{2}) c o s (\frac{a - b}{2})

c o s (\frac{π}{2^{n + 2}}) c o s (\frac{4 x - π}{2^{n + 2}})

\underset{n = 0}{\prod^{N}} c o s (\frac{a}{2^{n}}) = \frac{1}{s i n (\frac{a}{2^{N}})} \underset{n = 0}{\prod^{N}} c o s (\frac{a}{2^{n}}) . s i n (\frac{a}{2^{N}})

\Rightarrow \underset{n = 0}{\prod^{N}} c o s (\frac{a}{2^{n}}) = \frac{s i n (2 a)}{2^{N + 1} s i n (\frac{a}{2^{N}})}

l i m N \to 0 = \frac{s i n (2 a)}{2 a}

\underset{n = 0}{\prod^{N}} c o s (\frac{π}{4 . 2^{n}}) = \frac{s i n (\frac{π}{2})}{2^{N + 1} s i n (\frac{π}{4 . 2^{N}})} \to \frac{2}{π}

\underset{n = 0}{\prod^{N}} c o s (\frac{4 x - π}{2^{n} . 2}) = \frac{s i n (\frac{4 x - π}{2})}{(\frac{4 x - π}{2})} = \frac{2 s i n (2 x - \frac{π}{2})}{4 x - π} = \frac{2 c o s (2 x)}{π - 4 x}

\Rightarrow w e g e t . \frac{2}{π} . \frac{2 c o s (2 x)}{π - 4 x} = \frac{4 c o s (2 x)}{π (π - 4 x)}

Commented by ~blr237~ last updated on 28/Feb/20

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