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I'll give a proof of the following expansion:

$\frac{\sin x}{x} = \prod_{i=1}^{\infty} \cos \frac{x}{2^i}$

${\sin x} = 2 \cos \frac{x}{2}\sin \frac{x}{2}$

${\sin x} = 2^2 \cos \frac{x}{2}\cos \frac{x}{4}\sin \frac{x}{4}$

$ {\sin x} = 2^3 \cos \frac{x}{2} \cos \frac{x}{4} \cos \frac{x}{8} \sin\frac{x}{8} $

$ {\sin x} = 2^k \cos \frac{x}{2} \cos \frac{x}{4} \cdots\cos \frac{x}{2^k} \sin\frac{x}{2^k} $

${\sin x} = 2^k \sin\frac{x}{2^k} \prod_{i=1}^{k} \cos \frac{x}{2^i} $

$ \frac{\sin x}{x} = \frac{\sin \displaystyle \frac{x}{2^k}}{\displaystyle \frac{x}{2^k}} \prod_{i=1}^{k} \cos \frac{x}{2^i} $

Let $k \to \infty$, then $\displaystyle \frac{x}{2^k} \to 0$

$ \frac{\sin x}{x} = \prod_{i=1}^{\infty} \cos \frac{x}{2^i} $

Is there any extra observation needed to make the proof complete?

I guess since $\cos \frac{x}{2^i} \to 1$

the convergence is not at stake.

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    @AndréNicolas please add an answer rather than answering it in comments. If you are worried about gaining 'undeserved' reputation, you can always mark it as community wiki. We don't want the Community user to keep bumping this inspite of this being answered etc. See this meta thread: http://meta.math.stackexchange.com/questions/1559/dealing-with-answers-in-comments – 2012-02-08

1 Answers 1

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Your expression $ \frac{\sin x}{x} = \frac{\sin \frac{x}{2^k}}{ \frac{x}{2^k}} \prod_{i=1}^{k} \cos \frac{x}{2^i} $ is correct. Maybe we should separate out the very special case $x=0$, and from then on assume that $x\ne 0$. For $x=0$, $\frac{\sin x}{x}$ is formally undefined, but it is natural to set it equal to $1$. Then the formula works. If we are feeling in a very formal mood, we should prove the correctness of your expression by induction on $k$. However, I think that would be overkill.

We want to find the limit of the expression on the right as $k\to \infty$. As you observed, $\frac{x}{2^k}\to 0$ as $k\to \infty$. That certainly does not need proof. However, it is necessary to observe that since $\lim_{t=0}\frac{\sin t}{t}=1,$ we have $\lim_{k\to\infty} \frac{\sin \frac{x}{2^k}}{ \frac{x}{2^k}}=1.$ There will unfortunately be some special cases that require special treatment. We deal first with the "general" case when $x$ is not an integer multiple of $\pi$. Then your expression can be rewritten as $\frac{\sin x}{x}\frac{\frac{x}{2^k}}{\sin\frac{x}{2^k}}=\prod_{i=1}^{k} \cos \frac{x}{2^i}.$ Since $\lim_{k\to\infty}\frac{\sin x}{x}\frac{\frac{x}{2^k}}{\sin\frac{x}{2^k}}$ exists and is equal to $\frac{\sin x}{x}$, we conclude that $\lim_{k\to\infty} \prod_{i=1}^{k} \cos \frac{x}{2^i}$ also exists and is equal to $\frac{\sin x}{x}$.

If $x\ne 0$ is an integer multiple of $\pi$, we have to choose $k$ large enough so that $\sin\frac{x}{2^k}\ne 0$, else when we rewrite your expression, we might be dividing by $0$. Minor point. For such $x\ne 0$, $\frac{\sin x}{x}=0$, and one of the cosines is $0$. So, in that case also, apart from a technicality discussed below, the formula looks correct.

Technical remark: In the formal definition of an infinite product, we say that $\prod_{i=1}^\infty a_i$ converges if $\lim_{k\to\infty}\prod_{i=1}^k a_i$ exists and is not equal to $0$. So technically when $x$ is an integer multiple of $\pi$, the infinite product does not converge!