Conhecida as fórmulas de transformação $produto-soma$, ou simplesmente $\text{Fórmulas de Werner}$, temos: \begin{matrix} p = \sin \underbrace{{\dfrac{\pi}{24}} \cdot \ \sin\overbrace{{\dfrac{5\pi}{24}} \cdot \sin}^{produto-soma}{\dfrac{7\pi}{24}} \cdot \sin}_{produto-soma}{\dfrac{11\pi}{24}} \\ \\ p = {{ \left( \dfrac{\cos{\dfrac{12\pi}{24}} \ - \ \cos{\dfrac{10\pi}{24}} }{-2} \right)}} \cdot {{ \left( \dfrac{\cos{\dfrac{12\pi}{24}} \ - \ \cos{\dfrac{2\pi}{24}} }{-2} \right)}} = \dfrac{1}{4} \cdot \left(\cos{\dfrac{\pi}{2}} - \cos{\dfrac{5\pi}{12}}\right) \cdot \left( \cos{\dfrac{\pi}{2}} - \cos{\dfrac{\pi}{12}}\right) \\ \\ p = \dfrac{1}{4} \cdot \underbrace{\cos{\dfrac{5\pi}{12}} \cdot \cos{\dfrac{\pi}{12}}}_{produto-soma} \\ \\ p = \dfrac{1}{4} \cdot {{ \left( \dfrac{\cos{\dfrac{6\pi}{12}} \ + \ \cos{\dfrac{4\pi}{12}} }{2} \right)}} = \dfrac{1}{8} \cdot \left(\cos{\dfrac{\pi}{2}} + \cos{\dfrac{\pi}{3}}\right) = \dfrac{1}{8} \cdot \cos{\dfrac{\pi}{3}} \\ \\ \fbox{$p = { \dfrac{1}{16}}$} \end{matrix}

Outra maneira de resolver : $p = \sin(\frac{\pi}{24}) \cdot \sin(\frac{5\pi}{24}) \cdot \sin(\frac{7\pi}{24}) \cdot \sin(\frac{11\pi}{24}) $ $= p = \sin(\frac{\pi}{24}) \cdot \cos(\frac{\pi}{24}) \cdot \sin(\frac{5\pi}{24}) \cdot \cos(\frac{5\pi}{24})$ $\therefore$ $4p = (2\sin(\frac{\pi}{24}) \cdot \cos(\frac{\pi}{24})) \cdot (2\sin(\frac{5\pi}{24}) \cdot \cos(\frac{5\pi}{24}))$ $= 4p = \sin(\frac{\pi}{12}) \cdot \sin(\frac{5\pi}{12})$ $= 4p = \sin(\frac{\pi}{12}) \cdot \cos(\frac{\pi}{12}) $ $\therefore$ $8p = 2\sin(\frac{\pi}{12}) \cdot \cos(\frac{\pi}{12})$ $= 8p = \sin(\frac{\pi}{6}) = 8p = \dfrac{1}{2}$ $\implies \boxed{p = \dfrac{1}{16}}$

Utilizaremos os números complexos como abordagem para resolver esse problema. Para fazermos essa abordagem , primeiramente precisamos transformar essa multiplicação de senos em uma multiplicação de cossenos. $p = \sin(\frac{\pi}{24}) \cdot \sin(\frac{5\pi}{24}) \cdot \sin(\frac{7\pi}{24}) \cdot \sin(\frac{11\pi}{24})$ $ = \sin( \frac{\pi}{2} - \frac{11\pi}{24}) \cdot \sin(\frac{\pi}{2} - \frac{7\pi}{24}) \cdot \sin(\frac{\pi}{2} - \frac{5\pi}{24}) \cdot \sin(\frac{\pi}{2} - \frac{\pi}{24}) $ $= p = \cos(\frac{\pi}{24}) \cdot \cos(\frac{5\pi}{24}) \cdot \cos(\frac{7\pi}{24}) \cdot \cos(\frac{11\pi}{24}) $ Seja $z = a +bi $ , $a,b \in \mathbb{R} $ , tal que $|z| = 1$ , sabe-se que $z^n + \dfrac{1}{z^n} = 2\cos(n \theta)$ , sendo que $n \in \mathbb{Z} $ e $\theta $ é o ângulo que o vetor que representa o número complexo $ z $ realiza com o eixo $x$ , fazendo $\theta = \dfrac{\pi}{24}$ temos que : $\cos(\frac{\pi}{24}) \cdot \cos(\frac{5\pi}{24}) \cdot \cos(\frac{7\pi}{24}) \cdot \cos(\frac{11\pi}{24}) $ $ = p = \cos(\theta) \cdot \cos(5 \theta) \cdot \cos(7 \theta) \cdot \cos(11 \theta) $ $ = \dfrac{1}{2}\left(z + \dfrac{1}{z} \right) \cdot \dfrac{1}{2}\left(z^5+ \dfrac{1}{z^5}\right) \cdot \dfrac{1}{2}\left( z^7 + \dfrac{1}{z^7}\right) \cdot \dfrac{1}{2}\left( z^{11} + \dfrac{1}{z^{11}}\right)$ $ = \dfrac{1}{16}\left(z + \dfrac{1}{z} \right) \left(z^5 + \dfrac{1}{z^5}\right) \left(z^7 + \dfrac{1}{z^7} \right) \left(z^{11} + \dfrac{1}{z^{11}}\right)$ $= p = \dfrac{1}{16}(z^2 + \dfrac{1}{z^2} + z^8 + \dfrac{1}{z^8} + z^{10} + \dfrac{1}{z^{10}} + z^{12} + \dfrac{1}{z^{12}} + z^{14} + \dfrac{1}{z^{14}} + z^{22} + \dfrac{1}{z^{22}} + z^{24} + \dfrac{1}{z^{24}} + 2)$ Perceba que Se $n = 12 \implies z^{12}+\dfrac{1}{z^{12}} = 0 \implies z^{24} = -1$ Se $n = 24 \implies z^{24}+\dfrac{1}{z^{24}} = -2 \implies \dfrac{1}{z^{24}} = -1$ Manipulando essas duas equações , podemos encontrar as seguintes igualdades: $\boxed{z^{22} = -\dfrac{1}{z^2}}$ ; $\boxed{z^2 = -\dfrac{1}{z^{22}} }$ ; $\boxed{z^{14} =- \dfrac{1}{z^{10}} }$ ; $\boxed{z^{10} = -\dfrac{1}{z^{14}}} $ $\therefore$ $p = \dfrac{1}{16}(z^2 + \dfrac{1}{z^2} + z^8 + \dfrac{1}{z^8} + z^{10} + \dfrac{1}{z^{10}} + z^{12} + \dfrac{1}{z^{12}} + z^{14} + \dfrac{1}{z^{14}} + z^{22} + \dfrac{1}{z^{22}} + z^{24} + \dfrac{1}{z^{24}} + 2)$ $ = p = \dfrac{1}{16}(-\dfrac{1}{z^{22}} + \dfrac{1}{z^2} + z^8 + \dfrac{1}{z^8}-\dfrac{1}{z^{14}} + \dfrac{1}{z^{10}} +0 - \dfrac{1}{z^{10}} + \dfrac{1}{z^{14}} -\dfrac{1}{z^2} + \dfrac{1}{z^{22}} -2 + 2)$ $=p = \dfrac{1}{16}(-\cancel{\dfrac{1}{z^{22}}} + \cancel{\dfrac{1}{z^2}} + z^8 + \dfrac{1}{z^8}-\cancel{\dfrac{1}{z^{14}}} + \cancel{\dfrac{1}{z^{10}}} - \cancel{\dfrac{1}{z^{10}}} + \cancel{\dfrac{1}{z^{14}}} -\cancel{\dfrac{1}{z^2}} + \cancel{\dfrac{1}{z^{22}}} )$ $ p = \dfrac{1}{16}(z^8 + \dfrac{1}{z^8})$ $ = p =\dfrac{1}{16}(2 \cdot \cos(\frac{8\pi}{24})) $ $= p =\dfrac{1}{16}(2 \cdot \cos(\frac{\pi}{3})) $ $= p =\dfrac{1}{16}(2 \cdot \dfrac{1}{2}) $ $ = \boxed{p = \dfrac{1}{16}}$