Here is this famous identity due to Euler.

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

This holds true for every $x \ne 0$.

Let's prove it.

Denote:

$$ S(n) = \prod_{k=1}^{n} \cos \left(\frac{x}{2^k} \right) \tag{2}$$

Multiplying the two sides by $sin{\frac{x}{2^n}}$ we sequentially get:

$$ sin{\frac{x}{2^n}} \cdot S(n) = sin{\frac{x}{2^{n}}} \cdot \prod_{k=1}^{n} \cos \left(\frac{x}{2^k} \right) $$

$$ sin{\frac{x}{2^n}} \cdot S(n) = sin{\frac{x}{2^{n-1}}} \cdot \frac{1}{2^1} \cdot \prod_{k=1}^{n-1} \cos \left(\frac{x}{2^k} \right) $$

$$ sin{\frac{x}{2^n}} \cdot S(n) = sin{\frac{x}{2^{n-2}}} \cdot \frac{1}{2^2} \cdot \prod_{k=1}^{n-2} \cos \left(\frac{x}{2^k} \right) $$

$$ ... $$

$$ sin{\frac{x}{2^n}} \cdot S(n) = sin{\frac{x}{2^{1}}} \cdot \frac{1}{2^{n-1}} \cdot \prod_{k=1}^{1} \cos \left(\frac{x}{2^k} \right) $$

The last one obviously gives us:

$$ sin{\frac{x}{2^n}} \cdot S(n) = \frac{1}{2^{n}} \cdot sin(x) \tag{2}$$

which can be easily reworked to:

$$\large{ \frac{sin{\frac{x}{2^n}}}{\frac{x}{2^n}} \cdot S(n) = \frac{sin(x)}{x}} \tag{3}$$

Now in $(3)$ when we take the limit as ${n\to\infty}$ while using that

$$\lim_{u \to 0} \frac{sin(u)}{u} = 1$$

we get equality $(1)$ which is what we wanted to prove.