The problem asks to prove that $16 \sin(\frac{\pi}{24}) \sin(\frac{5\pi}{24}) \sin(\frac{7\pi}{24}) \sin(\frac{11\pi}{24}) = 1$.

1. Problem Description

The problem asks to prove that $16 \sin(\frac{\pi}{24}) \sin(\frac{5\pi}{24}) \sin(\frac{7\pi}{24}) \sin(\frac{11\pi}{24}) = 1$.

2. Solution Steps

We need to prove that $16 \sin(\frac{\pi}{24}) \sin(\frac{5\pi}{24}) \sin(\frac{7\pi}{24}) \sin(\frac{11\pi}{24}) = 1$.

First, we can rewrite $\sin(\frac{11\pi}{24})$ as $\sin(\pi - \frac{13\pi}{24}) = \sin(\frac{13\pi}{24})$. Also, $\sin(\frac{13\pi}{24}) = \sin(\frac{\pi}{2} + \frac{\pi}{24}) = \cos(\frac{\pi}{24})$.

So, the expression becomes:

$16 \sin(\frac{\pi}{24}) \sin(\frac{5\pi}{24}) \sin(\frac{7\pi}{24}) \cos(\frac{\pi}{24})$

$= 16 \sin(\frac{\pi}{24}) \cos(\frac{\pi}{24}) \sin(\frac{5\pi}{24}) \sin(\frac{7\pi}{24})$

$= 8 (2 \sin(\frac{\pi}{24}) \cos(\frac{\pi}{24})) \sin(\frac{5\pi}{24}) \sin(\frac{7\pi}{24})$

We use the formula $2 \sin x \cos x = \sin 2x$. Then:

$8 \sin(\frac{2\pi}{24}) \sin(\frac{5\pi}{24}) \sin(\frac{7\pi}{24})$

$= 8 \sin(\frac{\pi}{12}) \sin(\frac{5\pi}{24}) \sin(\frac{7\pi}{24})$

$\sin(\frac{\pi}{12}) = \sin(15^\circ) = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ = \frac{\sqrt{2}}{2} \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \frac{1}{2} = \frac{\sqrt{6} - \sqrt{2}}{4}$

Also, $\sin(\frac{5\pi}{24}) = \sin(37.5^\circ)$ and $\sin(\frac{7\pi}{24}) = \sin(52.5^\circ)$.

Using the product-to-sum formula, we have:

$\sin a \sin b = \frac{1}{2} [\cos(a-b) - \cos(a+b)]$

$\sin(\frac{5\pi}{24}) \sin(\frac{7\pi}{24}) = \frac{1}{2} [\cos(\frac{2\pi}{24}) - \cos(\frac{12\pi}{24})] = \frac{1}{2} [\cos(\frac{\pi}{12}) - \cos(\frac{\pi}{2})] = \frac{1}{2} \cos(\frac{\pi}{12})$

Since $\cos(\frac{\pi}{12}) = \cos(15^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4}$, we have

$\sin(\frac{5\pi}{24}) \sin(\frac{7\pi}{24}) = \frac{1}{2} (\frac{\sqrt{6} + \sqrt{2}}{4}) = \frac{\sqrt{6} + \sqrt{2}}{8}$

Therefore,

$8 \sin(\frac{\pi}{12}) \sin(\frac{5\pi}{24}) \sin(\frac{7\pi}{24}) = 8 (\frac{\sqrt{6} - \sqrt{2}}{4}) (\frac{\sqrt{6} + \sqrt{2}}{8}) = 8 (\frac{6-2}{32}) = 8 (\frac{4}{32}) = 8 (\frac{1}{8}) = 1$

3. Final Answer

$16 \sin(\frac{\pi}{24}) \sin(\frac{5\pi}{24}) \sin(\frac{7\pi}{24}) \sin(\frac{11\pi}{24}) = 1$