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Question Number 26999 by abdo imad last updated on 01/Jan/18

c a l c u l a t e \prod_{k = 1}^{n} c o s (\frac{a}{2^{k}}) a n d 0 < a < π t h e n f i n d t h e v a l u e o f

{l i m}_{n - >\propto} \sum_{k = 1}^{n} l n (c o s (\frac{a}{2^{k}})) .

Answered by prakash jain last updated on 01/Jan/18

\underset{k = 1}{\prod^{n}} \cos (\frac{a}{2^{k}})

= \frac{1}{\sin (\frac{a}{2^{n}})} \sin (\frac{a}{2^{n}}) \underset{k = 1}{\prod^{n}} \cos (\frac{a}{2^{k}})

= \frac{\sin a}{2^{n} \sin (\frac{a}{2^{n}})}

\lim_{n \to \infty} \frac{\sin a}{2^{n} \sin (\frac{a}{2^{n}})}

\lim_{n \to \infty} \frac{\sin a}{\frac{\sin (\frac{a}{2^{n}})}{\frac{a}{2^{7}}} \cdot a} = \frac{\sin a}{a}