Question 4 4.1. Given \( \sin \alpha=-\frac{8}{17^{\prime}} \), 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. \( \sin \underline{\alpha} \)

We are given \[ \sin \alpha = -\frac{8}{17}, \] with \[ 0^\circ < \alpha < 270^\circ. \] Since the sine is negative, and \(\alpha\) is between \(0^\circ\) and \(270^\circ\), the angle must lie in the third quadrant (because in the first quadrant \(\sin \alpha>0\) and in the second quadrant \(\sin \alpha>0\) as well). In the third quadrant both sine and cosine are negative. Using the Pythagorean identity \[ \sin^2 \alpha + \cos^2 \alpha = 1, \] we find \(\cos \alpha\). Compute \[ \sin^2 \alpha = \left(-\frac{8}{17}\right)^2 = \frac{64}{289}. \] Thus \[ \cos^2 \alpha = 1 - \frac{64}{289} = \frac{289-64}{289} = \frac{225}{289}. \] Since \(\alpha\) is in the third quadrant, \(\cos \alpha\) is negative. Hence, \[ \cos \alpha = -\frac{15}{17}. \] --- \(\mathbf{4.1.1.\; \tan \alpha}\) We use the definition: \[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha}. \] Substitute the values: \[ \tan \alpha = \frac{-\frac{8}{17}}{-\frac{15}{17}} = \frac{8}{15}. \] --- \(\mathbf{4.1.2.\; \cos 2\alpha}\) There are several formulas for \(\cos 2\alpha\). We may use either: \[ \cos 2\alpha = 2\cos^2 \alpha - 1 \quad \text{or} \quad \cos 2\alpha = 1 - 2\sin^2 \alpha. \] Using the cosine form: \[ \cos 2\alpha = 2\left(-\frac{15}{17}\right)^2 - 1 = 2\left(\frac{225}{289}\right) - 1 = \frac{450}{289} - 1 = \frac{450-289}{289} = \frac{161}{289}. \] --- \(\mathbf{4.1.3.\; \sin \underline{\alpha}}\) Here the notation \(\sin \underline{\alpha}\) is interpreted as the sine of the \emph{reference angle} for \(\alpha\). In the third quadrant, the reference angle, which we may denote as \(\alpha_r\), is defined by the absolute value of the sine: \[ \sin \alpha_r = \left|\sin \alpha\right| = \frac{8}{17}. \] --- Final Answers: \[ \tan \alpha = \frac{8}{15}, \] \[ \cos 2\alpha = \frac{161}{289}, \] \[ \sin \underline{\alpha} = \frac{8}{17}. \]