(b) If \( 2 \cos \alpha+1=0 \) and \( \sin \alpha<0 \), determine without the use of a calculator and with the aid of a diagram the value of: \( \begin{array}{lll}\text { (1) } \tan \alpha-\sin \alpha & \text { (2) } 2 \sin ^{2} \alpha-\cos ^{2} \alpha\end{array} \)

To solve for \( \alpha \), we start with the equation \( 2 \cos \alpha + 1 = 0 \). Rearranging gives us \( 2 \cos \alpha = -1 \), leading to \( \cos \alpha = -\frac{1}{2} \). From the unit circle, \( \cos \alpha = -\frac{1}{2} \) corresponds to angles in the second and third quadrants. However, given that \( \sin \alpha < 0 \), we are in the third quadrant. Therefore, \( \alpha = \frac{4\pi}{3} \) or \( 240^\circ \). Now we can find the values of \( \sin \alpha \) and \( \tan \alpha \): - In the third quadrant, \( \sin \alpha = -\frac{\sqrt{3}}{2} \) and \( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = \sqrt{3} \). Now, let's calculate the two expressions. 1) \( \tan \alpha - \sin \alpha \): \[ \tan \alpha - \sin \alpha = \sqrt{3} - \left(-\frac{\sqrt{3}}{2}\right) = \sqrt{3} + \frac{\sqrt{3}}{2} = \frac{2\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}. \] 2) \( 2 \sin^2 \alpha - \cos^2 \alpha \): To find \( \sin^2 \alpha \) and \( \cos^2 \alpha \): - \( \sin^2 \alpha = \left(-\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \). - \( \cos^2 \alpha = \left(-\frac{1}{2}\right)^2 = \frac{1}{4} \). Now substitute: \[ 2 \sin^2 \alpha - \cos^2 \alpha = 2 \left(\frac{3}{4}\right) - \frac{1}{4} = \frac{6}{4} - \frac{1}{4} = \frac{5}{4}. \] Final answers: 1) \( \tan \alpha - \sin \alpha = \frac{3\sqrt{3}}{2} \) 2) \( 2 \sin^2 \alpha - \cos^2 \alpha = \frac{5}{4} \)