algebra precalculus - Calculate $sqrt{frac{1}{2}} times sqrt{frac{1}{2} + frac{1}{2}sqrt{frac{1}{2}}} times ldots$

$$\sqrt{\frac{1}{2}} \times \sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2}}} \times \sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2}+ \frac{1}{2}\sqrt{\frac{1}{2}}}} \times\ldots$$

I already know a way to calculate it:

With $\cos{\frac{\pi}{4}} = \sqrt{\frac{1}{2}}$ and denote $\frac{\pi}{4} = x$. Observe that:

$$\sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2}}} = \sqrt{\frac{1+\cos{x}}{2}} = \cos{\frac{x}{2}} \\ \sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2}+ \frac{1}{2}\sqrt{\frac{1}{2}}}} = \cos\frac{x}{4}$$

Thus it becomes

$$\begin{align} P(n) &= \cos{\frac{x}{2^n}}\cos{\frac{x}{2^{n-1}}} \cdots \cos{x} \\ &= \frac{2\sin{\frac{x}{2^{n-1}}}\cos{\frac{x}{2^{n-1}}}\cos{\frac{x}{2^{n-2}}} \cdots\cos{\frac{x}{2}}}{2\sin{\frac{x}{2^{n-1}}}} \end{align}$$

Taking in to account that $2\sin{x}\cos{x} = \sin{2x}$, we have

$$\begin{align} P(n) &= \lim_{n \to \infty} \frac{\sin{2x}}{2^n\sin{\frac{x}{2^{n-1}}}} \\ &= \lim_{n\rightarrow \infty} \frac{\sin{2x}}{2x} \\ &= \frac{2\sin\frac{\pi}{2}}{\pi} \\ &= \boxed{\frac{2}{\pi}} \end{align}$$

Now, I'm looking for another solution, please comment on.