Question 4 4.1. Given \( \sin \alpha=-\frac{8}{17} \), where \( 0^{\circ}<\alpha<270^{\circ} \), determine with the aid of a sketch, the value of the following: 4.1.1. \( \tan \alpha \) 4.1.2. \( \cos 2 \alpha \) 4.1.3. \( \sin \frac{\alpha}{2} \) 4.2. Simplify the following trigonometric expression, leaving your answer to only one trigonometric ratio: \( \left[\sin \left(x-1980^{\circ}\right)-\cos (-x)\right]^{2} \)

To tackle \( \tan \alpha \), we first note that the angle \( \alpha \) lies in the third quadrant since \( \sin \alpha = -\frac{8}{17} \). Using the Pythagorean identity, we find \( \cos \alpha \). Thus, \( \cos^2 \alpha = 1 - \sin^2 \alpha = 1 - (-\frac{8}{17})^2 = \frac{15}{17} \), leading to \( \cos \alpha = -\frac{\sqrt{15}}{17} \) (since cosine is also negative in the third quadrant). Therefore, \( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{-\frac{8}{17}}{-\frac{\sqrt{15}}{17}} = \frac{8}{\sqrt{15}} \). Using the double angle formula, \( \cos 2\alpha = 2\cos^2 \alpha - 1 \) gives us \( \cos 2\alpha = 2 \left(-\frac{\sqrt{15}}{17}\right)^2 - 1 = \frac{30}{289} - 1 = \frac{-259}{289} \). Now for \( \sin \frac{\alpha}{2} \), we apply the half-angle formula: \( \sin \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{2}} \). This leads to \( \sin \frac{\alpha}{2} = \sqrt{\frac{1 + \frac{\sqrt{15}}{17}}{2}} = \sqrt{\frac{17 + \sqrt{15}}{34}} \). For the expression \( \left[\sin (x-1980^{\circ})-\cos (-x)\right]^{2} \), we first simplify \( \sin (x-1980^{\circ}) \) since \( 1980^{\circ} \) is equivalent to \( 540^{\circ} \) (due to the periodicity of sine and cosine). Thus, we find \( \sin (x-540^{\circ}) = \sin (x) \). As \( \cos(-x) = \cos(x) \), we rewrite our expression as \( [\sin(x) - \cos(x)]^2 \). The final simplified result is \( \sin^2(x) - 2\sin(x)\cos(x) + \cos^2(x) = 1 - \sin(2x) \).