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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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@Peter: I've deleted the earlier version of your question. Please use the "edit" feature (the button is on the lower left of your post, just below the tags) to make changes. – Zev Chonoles Feb 8 '12 at 17:37
@ZevChonoles I did use it. What happened? – Pedro Tamaroff Feb 8 '12 at 17:40
There was a earlier copy of your question posted a few minutes before this one here; I'll undelete it for a bit so you can see. (I'm sure it was an honest mistake - don't worry about it.) – Zev Chonoles Feb 8 '12 at 17:44
You're wrong to say that 'convergence is not at stake' just because the terms of the infinite product go to 1 - just as you can't say that a series converges just because its terms go to 0. (And in fact, the example is essentially the same - consider $\Pi_{n=0}^\infty(1+1/n)$; the terms of the product clearly go to 1, but the product diverges.) Fortunately, you've already essentially shown the product converges by showing that the partial products approach a limit - I would just make this more explicit. – Steven Stadnicki Feb 8 '12 at 18:59
@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:… – Aryabhata Feb 8 '12 at 19:55
up vote 5 down vote accepted

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!

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