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Question Number 192985 by MM42 last updated on 01/Jun/23

p r o v e i t :

{l i m}_{n \to \infty} \underset{i = 1}{\prod^{n}} c o s \frac{θ}{2^{i}} = \frac{s i n θ}{θ}

t h e n s h o w :

{i m}_{n \to \infty} c o s \frac{π}{4} c o s \frac{π}{8} \dots c o s \frac{π}{2^{n + 1}} = \frac{2}{π}

Answered by witcher3 last updated on 02/Jun/23

\cos (\frac{θ}{2^{i}}) = \frac{\sin (\frac{θ}{2^{i - 1}})}{2 \sin (\frac{θ}{2^{i}})}

A_{n} = \underset{i = 1}{\prod^{n}} \cos (\frac{θ}{2^{i}}) = \underset{1}{\prod^{n}} \frac{\sin (\frac{θ}{2^{i - 1}})}{2 \sin (\frac{θ}{2^{i}})} = \frac{\sin (θ)}{2^{n} \sin (\frac{θ}{2^{n}})}

Double subscripts: use braces to clarify

(2) θ = \frac{π}{2}

Commented by MM42 last updated on 02/Jun/23

v e r y g o o d

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