An Infinite Trigonometric Product

$\begin{aligned} f(x) & = \prod_{n=1}^\infty \cos \frac x{2^n} \\ & = \lim_{n \to \infty} \cos \frac x{2} \times \cos \frac x{2^2} \times \cos \frac x{2^3} \times \cdots \times \cos \frac x{2^{n-2}} \times \cos \frac x{2^{n-1}} \times \cos \frac x{2^n} \\ & = \lim_{n \to \infty} \cos \frac x{2} \times \cos \frac x{2^2} \times \cos \frac x{2^3} \times \cdots \times \cos \frac x{2^{n-2}} \times \cos \frac x{2^{n-1}} \times \cos \frac x{2^n} \times \frac {\sin \frac x{2^n}}{\sin \frac x{2^n}} \\ & = \lim_{n \to \infty} \cos \frac x{2} \times \cos \frac x{2^2} \times \cos \frac x{2^3} \times \cdots \times \cos \frac x{2^{n-2}} \times \cos \frac x{2^{n-1}} \times \frac {\sin \frac x{2^{n-1}}}{2 \sin \frac x{2^n}} \\ & = \lim_{n \to \infty} \cos \frac x{2} \times \cos \frac x{2^2} \times \cos \frac x{2^3} \times \cdots \times \cos \frac x{2^{n-2}} \times \frac {\sin \frac x{2^{n-2}}}{2^2 \sin \frac x{2^n}} \\ & = \lim_{n \to \infty} \cos \frac x{2} \times \cos \frac x{2^2} \times \cos \frac x{2^3} \times \cdots \times \frac {\sin \frac x{2^{n-3}}}{2^3 \sin \frac x{2^n}} \\ & = \lim_{n \to \infty} \cos \frac x{2} \times \frac {\sin \frac x{2}}{2^{n-1} \sin \frac x{2^n}} \\ & = \lim_{n \to \infty} \frac {\sin x}{2^n \sin \frac x{2^n}} \\ & = \lim_{n \to \infty} \frac {\sin x}{\frac {x \sin \frac x{2^n}}{\frac x{2^n}}} \\ & = \frac {\sin x}x \end{aligned}$

Therefore, $f(1) = \dfrac {\sin 1}1 \approx \boxed{0.841}$ .

Let $P_N = \prod_{i = 1}^N \cos \frac{x}{2^n} = \frac{\sin x}{2^N \sin\frac{x}{2^N}}.$ Taking the limit as $N \to \infty$ , we get $\frac{\sin x}{x}$ , giving the answer $\boxed{\sin 1}$ .