Anybody know a proof of $\prod_{n=1}^\infty\cos(x/2^n)=\sin x/x$.

Question

This is actually an exercise from Apostol's Mathematical Analysis. Ch. 8 Ex 42. which asks to find all real values $x$ for which $\prod_{n=1}^\infty \cos\left(\large\frac{x}{2^n}\right)$ converges. I've shown that the product converges for all $x$. The problem then asks to find what values the product converges to. By playing around with Wolfram Alpha, I found that $$\large\prod_{n=1}^\infty\cos\left(\frac{x}{2^n}\right)=\frac{\sin (x)}{x}.$$

I can't figure out how to prove this.

Answer 1

Using the trig identity

$$\sin (2t) = 2\sin (t) \cos (t),$$

we have

$$\prod_{n = 1}^N \cos(x/2^n) = \prod_{n = 1}^N \frac{\sin(x/2^{n-1})}{2\sin(x/2^n)} = \frac{\sin(x)}{2^N\sin(x/2^N)} = \frac{\sin x}{x}\cdot \frac{x/2^N}{\sin(x/2^N)}$$

Take the limit as $N \to \infty$ and use the fact $\lim_{t\to 0} \frac{\sin t}{t} = 1$ to obtain the result.

Answer 2

Hint

$$\cos(x/2^n)=\frac12\frac{\sin(x/2^{n-1})}{\sin(x/2^n)}$$ and telescope.