2. Sendo \( \operatorname{sen} x+\cos x=\frac{\sqrt{5}}{4} \), calcule: \( \begin{array}{ll}\text { a) } \operatorname{sen} x \cdot \cos x & \text { b) } \operatorname{sen}^{3} x+\cos ^{3} x\end{array} \)

Para resolver \( \operatorname{sen} x + \cos x = \frac{\sqrt{5}}{4} \), vamos primeiro elevar ambos os lados ao quadrado: \[ (\operatorname{sen} x + \cos x)^2 = \left(\frac{\sqrt{5}}{4}\right)^2 \] Isso dá: \[ \operatorname{sen}^2 x + 2\operatorname{sen} x \cos x + \cos^2 x = \frac{5}{16} \] Sabendo que \( \operatorname{sen}^2 x + \cos^2 x = 1 \), temos: \[ 1 + 2\operatorname{sen} x \cos x = \frac{5}{16} \] Subtraindo 1 de ambos os lados: \[ 2\operatorname{sen} x \cos x = \frac{5}{16} - 1 = \frac{5}{16} - \frac{16}{16} = \frac{-11}{16} \] Portanto, isolamos \( \operatorname{sen} x \cdot \cos x \): \[ \operatorname{sen} x \cdot \cos x = \frac{-11}{32} \] Agora para o segundo item: Para calcular \( \operatorname{sen}^3 x + \cos^3 x \), podemos usar a fórmula de soma de cubos: \[ \operatorname{sen}^3 x + \cos^3 x = (\operatorname{sen} x + \cos x)(\operatorname{sen}^2 x - \operatorname{sen} x \cos x + \cos^2 x) \] Já sabemos que \( \operatorname{sen} x + \cos x = \frac{\sqrt{5}}{4} \) e que \( \operatorname{sen}^2 x + \cos^2 x = 1 \), então: \[ \operatorname{sen}^2 x - \operatorname{sen} x \cos x + \cos^2 x = 1 - \operatorname{sen} x \cos x \] Substituindo \( \operatorname{sen} x \cdot \cos x = \frac{-11}{32} \): \[ 1 - \operatorname{sen} x \cos x = 1 - \frac{-11}{32} = 1 + \frac{11}{32} = \frac{32 + 11}{32} = \frac{43}{32} \] Agora, substituímos tudo na fórmula de soma de cubos: \[ \operatorname{sen}^3 x + \cos^3 x = \left(\frac{\sqrt{5}}{4}\right)\left(\frac{43}{32}\right) = \frac{43\sqrt{5}}{128} \] Portanto, as respostas são: a) \( \operatorname{sen} x \cdot \cos x = \frac{-11}{32} \) b) \( \operatorname{sen}^3 x + \cos^3 x = \frac{43\sqrt{5}}{128} \)