Activity 4 If \( \cos \beta=\frac{p}{\sqrt{5}} \) where \( p<0 \) and \( \beta \in\left[180^{\circ} ; 360^{\circ}\right] \), determine, using a diagram, an expression in terms of \( p \) for: \( \begin{array}{l}4.1 .1 \quad \tan \beta \\ 4.1 .2 \quad \cos 2 \beta\end{array} \)

To determine the expressions for \( \tan \beta \) and \( \cos 2\beta \) given that \( \cos \beta = \frac{p}{\sqrt{5}} \) where \( p < 0 \), we can start by visualizing the unit circle. Since \( \beta \) is in the interval \( [180^\circ, 360^\circ] \), it means that \( \beta \) is in the third or fourth quadrant, where the cosine value is negative. This means \( \cos \beta = \frac{p}{\sqrt{5}} < 0 \). Using the Pythagorean identity \( \sin^2 \beta + \cos^2 \beta = 1 \), we can find \( \sin \beta \). First, we can express \( \sin^2 \beta \): \[ \sin^2 \beta = 1 - \cos^2 \beta = 1 - \left(\frac{p}{\sqrt{5}}\right)^2 = 1 - \frac{p^2}{5} = \frac{5 - p^2}{5} \] Since \( \beta \in [180^\circ, 360^\circ] \), \( \sin \beta < 0 \). Thus, \( \sin \beta = -\sqrt{\frac{5 - p^2}{5}} = -\frac{\sqrt{5 - p^2}}{\sqrt{5}} \). 4.1.1 For \( \tan \beta \): \[ \tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{-\frac{\sqrt{5 - p^2}}{\sqrt{5}}}{\frac{p}{\sqrt{5}}} = -\frac{\sqrt{5 - p^2}}{p} \] 4.1.2 For \( \cos 2\beta \): Using the double angle formula \( \cos 2\beta = 2\cos^2 \beta - 1 \): \[ \cos 2\beta = 2\left(\frac{p}{\sqrt{5}}\right)^2 - 1 = 2\cdot\frac{p^2}{5} - 1 = \frac{2p^2}{5} - 1 \] Thus, the expressions in terms of \( p \) are: \[ \text{4.1.1 } \tan \beta = -\frac{\sqrt{5 - p^2}}{p} \] \[ \text{4.1.2 } \cos 2\beta = \frac{2p^2}{5} - 1 \]