# Limit n tends to infinity

How can i solve this: $$\lim_{n\to\infty} \cos(1)\cos(0.5)\cos(0.25)\ldots \cos(1/2^n)$$

I tried using comlex numbers and logarithms but did'nt work out.Can anyone help please.

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hint: multiply $sin \dfrac{1}{2^n}$

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Brilliant hint. Why this post was reported as low quality is beyond me... – Lost1 Dec 26 '13 at 13:50

Note that $$\sin(x) = 2 \sin(x/2) \cos(x/2) = 2^2\sin(x/4) \cos(x/2) \cos(x/4) = \ldots = 2^n \sin(x/2^n)\prod_{k=1}^n \cos(x/2^k)$$ Hence, $$\lim_{n\to\infty} \prod_{k=1}^{n} \cos(x/2^k) = \lim_{n\to \infty} \frac{\sin(x)}{2^n \sin(x/2^n)} = \frac{\sin(x)}{x} \lim_{n\to\infty} \frac{x/2^n}{\sin(x/2^n)}$$ Now use the fact that $$\lim_{z \to 0}\frac{\sin(z)}{z} = 1$$ Now plug in $x=1$.

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so answer is sin1 ? – Devgeet Patel Dec 26 '13 at 13:28
very cool answer – Don Larynx Dec 26 '13 at 13:35

We have the infinite product expansion

$$\frac{\sin x}{x}=\prod _{k=1}^\infty \cos\left(\frac{x}{2^k}\right)$$

This result was first found by Francois Viète in 1593. For a proof, see also this math.se question.

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