1. **Problema**: Calcular os valores das funções trigonométricas dadas: (a) $\sin\left(\frac{4\pi}{6}\right)$ (b) $\cos\left(\frac{8\pi}{6}\right)$ (c) $\tan\left(\frac{5\pi}{4}\right)$ (d) $\cos\left(\frac{\pi}{12}\right)$ (e) $\sin\left(\frac{\pi}{12}\right)$ (f) $\sin\left(\frac{\pi}{8}\right)$. 2. Para facilitar, vamos expressar os ângulos e usar identidades e valores conhecidos. **(a)** $\sin\left(\frac{4\pi}{6}\right) = \sin\left(\frac{2\pi}{3}\right)$. Sabemos que $\sin\left(\frac{2\pi}{3}\right) = \sin\left(\pi - \frac{\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$. **(b)** $\cos\left(\frac{8\pi}{6}\right) = \cos\left(\frac{4\pi}{3}\right)$. Isso é no terceiro quadrante onde o cosseno é negativo. $\cos\left(\frac{4\pi}{3}\right) = \cos\left(\pi + \frac{\pi}{3}\right) = - \cos\left(\frac{\pi}{3}\right) = -\frac{1}{2}$. **(c)** $\tan\left(\frac{5\pi}{4}\right)$. Este ângulo está no terceiro quadrante onde a tangente é positiva. $\tan\left(\frac{5\pi}{4}\right) = \tan\left(\pi + \frac{\pi}{4}\right) = \tan\left(\frac{\pi}{4}\right) = 1$. **(d)** Para $\cos\left(\frac{\pi}{12}\right)$ usamos a fórmula da soma: $\frac{\pi}{12} = \frac{\pi}{3} - \frac{\pi}{4}$. $$\cos\left(\frac{\pi}{12}\right) = \cos\left(\frac{\pi}{3} - \frac{\pi}{4}\right) = \cos\frac{\pi}{3}\cos\frac{\pi}{4} + \sin\frac{\pi}{3}\sin\frac{\pi}{4}.$$ Sabemos: $\cos\frac{\pi}{3} = \frac{1}{2}$, $\cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}$, $\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$, $\sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}$. Substituindo: $$\cos\left(\frac{\pi}{12}\right) = \frac{1}{2} \cdot \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}.$$ **(e)** Para $\sin\left(\frac{\pi}{12}\right) = \sin\left(\frac{\pi}{3} - \frac{\pi}{4}\right)$. Usamos: $$\sin(a - b) = \sin a \cos b - \cos a \sin b,$$ portanto: $$\sin\left(\frac{\pi}{12}\right) = \sin\frac{\pi}{3}\cos\frac{\pi}{4} - \cos\frac{\pi}{3}\sin\frac{\pi}{4} = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} - \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}.$$ **(f)** Para $\sin\left(\frac{\pi}{8}\right)$ usamos a fórmula do ângulo metade: $$\sin\frac{x}{2} = \sqrt{\frac{1 - \cos x}{2}}.$$ Aqui, $x = \frac{\pi}{4}$ pois $\frac{\pi}{8} = \frac{\pi}{4} / 2$. Sabemos $\cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}$. Logo: $$\sin\left(\frac{\pi}{8}\right) = \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}} = \sqrt{\frac{2 - \sqrt{2}}{4}} = \frac{\sqrt{2 - \sqrt{2}}}{2}.$$ 3. **Respostas finais:** \begin{align*} (a)\quad & \sin\left(\frac{4\pi}{6}\right) = \frac{\sqrt{3}}{2} \\ (b)\quad & \cos\left(\frac{8\pi}{6}\right) = -\frac{1}{2} \\ (c)\quad & \tan\left(\frac{5\pi}{4}\right) = 1 \\ (d)\quad & \cos\left(\frac{\pi}{12}\right) = \frac{\sqrt{6} + \sqrt{2}}{4} \\ (e)\quad & \sin\left(\frac{\pi}{12}\right) = \frac{\sqrt{6} - \sqrt{2}}{4} \\ (f)\quad & \sin\left(\frac{\pi}{8}\right) = \frac{\sqrt{2 - \sqrt{2}}}{2} \end{align*} Todos os valores foram obtidos usando identidades trigonométricas e valores conhecidos para ângulos especiais.