Question

The value of \[ \frac{\pi}{2} \lim_{n \to \infty} \cos\!\left(\frac{\pi}{4}\right) \cos\!\left(\frac{\pi}{8}\right) \cos\!\left(\frac{\pi}{16}\right) \cdots \cos\!\left(\frac{\pi}{2^{n+1}}\right) \] is _________.

Hint:

Use the trigonometric product formula \(\sin x = x \prod_{k=1}^{\infty} \cos(x / 2^k)\) for problems involving infinite cosine products.

Answer 1

Correct Answer: 1

Step 1: Use the known infinite product identity.
\[ \sin x = x \prod_{k=1}^{\infty} \cos\!\left(\frac{x}{2^k}\right). \] Let \(x = \frac{\pi}{2}.\)
Step 2: Substitute and simplify.
\[ \sin\!\left(\frac{\pi}{2}\right) = \frac{\pi}{2} \prod_{k=1}^{\infty} \cos\!\left(\frac{\pi}{2^{k+1}}\right). \] \[ 1 = \frac{\pi}{2} \prod_{k=1}^{\infty} \cos\!\left(\frac{\pi}{2^{k+1}}\right). \]
Step 3: Rearranged result.
\[ \frac{\pi}{2} \prod_{k=1}^{\infty} \cos\!\left(\frac{\pi}{2^{k+1}}\right) = 1. \]
Step 4: Conclusion.
Hence, the required value is \(\boxed{1}.\)