\( \tan 2 \alpha \) \( \tan 2 \theta=-\sqrt{5} \) with \( 0^{\circ}<2 \theta<180^{\circ} \), determine the value of the following without the ule calculator and with the aid of a diagram: \( \begin{array}{lll}\sin 2 \theta & \text { (2) } \sin \theta \cos \theta & \text { (3) } 2 \cos ^{2} \theta \\ \cos \theta & & \end{array} \)

To find \( \sin 2\theta \), \( \sin \theta \cos \theta \), \( 2 \cos^2 \theta \), and \( \cos \theta \) given that \( \tan 2\theta = -\sqrt{5} \), we start by acknowledging that since \( \tan 2\theta \) is negative and the angle \( 2\theta \) is in the range \( 0^{\circ} < 2\theta < 180^{\circ} \), this indicates that \( 2\theta \) must be in the second quadrant (where tangent is negative). From \( \tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} = -\sqrt{5} \), we can draw a right triangle where the opposite side is \(-\sqrt{5}\) and the adjacent side is \(1\). The hypotenuse then is \(\sqrt{1^2 + (-\sqrt{5})^2} = \sqrt{6}\). This gives us: - \( \sin 2\theta = \frac{-\sqrt{5}}{\sqrt{6}} = -\frac{\sqrt{30}}{6} \) - \( \cos 2\theta = \frac{1}{\sqrt{6}} = \frac{\sqrt{6}}{6} \) Using the double angle identities, we find the values of \( \sin \theta \) and \( \cos \theta \): 1. \( \sin 2\theta = 2 \sin \theta \cos \theta \) implies \( 2 \sin \theta \cos \theta = -\frac{\sqrt{30}}{6} \) 2. Therefore, \( \sin \theta \cos \theta = -\frac{\sqrt{30}}{12} \) 3. The cosine can also be determined from \( \cos 2\theta = 2\cos^2 \theta - 1 \), leading to \( 2\cos^2 \theta = \frac{\sqrt{6}}{6} + 1 \) 4. Solving gives \( \cos^2 \theta = \frac{3+\sqrt{6}}{12} \) Thus, you will have: - \( \sin 2\theta = -\frac{\sqrt{30}}{6} \) - \( \sin \theta \cos \theta = -\frac{\sqrt{30}}{12} \) - \( 2 \cos^2 \theta = \frac{3+\sqrt{6}}{6} \) - \( \cos \theta = \sqrt{\frac{3+\sqrt{6}}{12}} \) These answers maintain the relationships and signs necessary based on the quadrant. Keep in mind that visual aids, like diagrams of the unit circle or right triangles, can drastically enhance the understanding of where these values lie on the coordinate plane. Drawing it out with angles and triangle shapes often makes complex trigonometric relations much easier to grasp!