• Resolvendo a (A) Veja que $\sin{\frac{\pi}{2n}}, n \in \mathbb{Z}^+$ está limitado ao primeiro quadrante, o que significa que $\sin{\frac{\pi}{2n}} > 0$ \begin{matrix} (\sin{\frac{\pi}{2n}})^2 = \frac{2 - \sqrt{3}}{4} = \frac{1}{2} - \frac{\sqrt{3}}{4} \\ \\ 2.(\sin{\frac{\pi}{2n}})^2 = 1 - \frac{\sqrt{3}}{2} \end{matrix} $\color{orangered}{Obs \ (1):}$ \begin{matrix} \cos{2x} = 1 - 2.\sin^2{x}\end{matrix} Logo: \begin{matrix} \cos{[2.(\frac{\pi}{2.n})]}= \frac{\sqrt{3}}{2} \\ \\ \cos{(\frac{\pi}{n})}= \frac{\sqrt{3}}{2} \ \ \rightarrow \ \ n=6 \end{matrix} • Resolvendo a (B) Já sabemos que $\sin{\frac{\pi}{12}} = \sqrt{\frac{2 - \sqrt{3}}{4}}$, então: $\color{orangered}{Obs \ (2):}$ \begin{matrix} \sin{2x}= 2.\sin{x}.cos{x}\end{matrix} Vejamos: \begin{matrix} \sin{\frac{\pi}{6}} =2.\sin{\frac{\pi}{12}}.\cos{\frac{\pi}{12}} \\ \\ \cos{\frac{\pi}{12}} = \frac{1}{2}.\sqrt{\frac{1}{2-\sqrt{3}}} \\ \\ \cos{\frac{\pi}{12}} = 1 - 2.\sin^2{\frac{\pi}{24}} \\ \\ \sin{\frac{\pi}{24}} = \frac{\sqrt{2 -\sqrt{2 + \sqrt{3}}}}{2} \end{matrix}