Questão 24 - Prova Modelo A - AFA 2026

$$\begin{gathered} \mathrm{senx} = \frac{\sqrt{5} - 1}{4} \Rightarrow \mathrm{cosx} = \sqrt{1 - {\mathrm{sen}}^2\mathrm{x} } = \frac{\sqrt{10 + 2\sqrt{5}}}{4} \\ \mathrm{tgx} = \frac{\mathrm{senx}}{\mathrm{cosx}} = \frac{\sqrt{5} - 1}{\sqrt{10 + 2\sqrt{5}}} \end{gathered}$$

$\mathrm{tg}2\mathrm{x} = \frac{2\mathrm{tgx}}{1 - {\mathrm{tg}}^2\mathrm{x}} = \frac{2 · {\displaystyle \frac{\sqrt{5} -1}{\sqrt{10 + 2\sqrt{5}}}}}{1 - {\left({\displaystyle \frac{\sqrt{5} - 1}{\sqrt{10 + 2\sqrt{5}}}}\right)}^2} = \left(\sqrt{5} - 1\right) · \frac{\sqrt{10 + 2\sqrt{5}}}{2(\sqrt{5} + 1)}$

 

$$\begin{gathered} \mathrm{tg}3\mathrm{x} = \frac{\mathrm{tgx} + \mathrm{tg}2\mathrm{x}}{1 - \mathrm{tgx} · \mathrm{tg}2\mathrm{x}} = \frac{{\displaystyle \frac{\sqrt{5} - 1}{\sqrt{10 + 2\sqrt{5}}} + \left(\sqrt{5} - 1\right) · \frac{\sqrt{10 + 2\sqrt{5}}}{2\left(\sqrt{5} + 1\right)}}}{1 - {\displaystyle \frac{\sqrt{5} - 1}{\sqrt{10 + 2\sqrt{5}}}} · \left(\sqrt{5} - 1\right) · {\displaystyle \frac{\sqrt{10 + 2\sqrt{5}}}{2\left(\sqrt{5} + 1\right)}} } = \\ =\frac{{\displaystyle \frac{2 · \left(\sqrt{5} - 1\right) · \left(\sqrt{5} + 1\right) + \left(\sqrt{5} - 1\right) · \left(10 + 2\sqrt{5}\right)}{2 · \left(\sqrt{5} + 1\right) · \sqrt{10 + 2\sqrt{5}}} }}{{\displaystyle \frac{2\sqrt{5} + 2 - \left(6 - 2\sqrt{5}\right)}{2 · (\sqrt{5} + 1)}}} = \\ = \frac{{\displaystyle \frac{8 + 8\sqrt{5}}{\sqrt{10 + 2\sqrt{5}}}}}{4 · \left(\sqrt{5} - 1\right)} \end{gathered}$$

$$\begin{gathered} \mathrm{tg}3\mathrm{x} = 2 · \frac{\sqrt{5} + 1}{\sqrt{5} - 1} · \frac{1}{\sqrt{10 + 2\sqrt{5}}} \Rightarrow \\ \Rightarrow \mathrm{cotg}3\mathrm{x} = \sqrt{10 + 2\sqrt{5}} · \frac{\sqrt{5} - 1}{2 · \left(\sqrt{5} + 1\right) } \Rightarrow \\ \Rightarrow {\mathrm{cotg}}^{2}3\mathrm{x} = \left(10 + 2\sqrt{5}\right) · \frac{{\left(\sqrt{5} - 1\right)}^2}{4 · {\left(\sqrt{5} + 1\right)}^2} \end{gathered}$$

Efetuando os cálculos, tem-se: 

$\boxed{{\mathrm{cotg}}^{2}3\mathrm{x} = 5 - 2\sqrt{5}}$