\( \tan 2 \alpha \) \( \tan 2 \theta=-\sqrt{5} \) 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 solve the problem, we start with the given condition: \[ \tan 2\theta = -\sqrt{5} \] ### Step 1: Determine the Quadrant for \(2\theta\) Since \(0^{\circ} < 2\theta < 180^{\circ}\), the angle \(2\theta\) must be in the second quadrant because the tangent is negative in the second quadrant. ### Step 2: Find \( \sin 2\theta \) and \( \cos 2\theta \) Using the identity for tangent: \[ \tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} \] Let \( \sin 2\theta = y \) and \( \cos 2\theta = x \). Then we have: \[ \frac{y}{x} = -\sqrt{5} \implies y = -\sqrt{5}x \] Since \(2\theta\) is in the second quadrant, we know that: - \(y = \sin 2\theta > 0\) - \(x = \cos 2\theta < 0\) Using the Pythagorean identity: \[ \sin^2 2\theta + \cos^2 2\theta = 1 \] Substituting \(y\) into the identity: \[ (-\sqrt{5}x)^2 + x^2 = 1 \] This simplifies to: \[ 5x^2 + x^2 = 1 \implies 6x^2 = 1 \implies x^2 = \frac{1}{6} \implies x = -\frac{1}{\sqrt{6}} \quad (\text{since } x < 0) \] Now substituting back to find \(y\): \[ y = -\sqrt{5}x = -\sqrt{5} \left(-\frac{1}{\sqrt{6}}\right) = \frac{\sqrt{5}}{\sqrt{6}} = \frac{\sqrt{30}}{6} \] ### Step 3: Calculate \( \sin 2\theta \), \( \sin \theta \cos \theta \), and \( 2 \cos^2 \theta \) 1. **Finding \( \sin 2\theta \)**: \[ \sin 2\theta = \frac{\sqrt{30}}{6} \] 2. **Finding \( \cos 2\theta \)**: \[ \cos 2\theta = -\frac{1}{\sqrt{6}} \] 3. **Using the double angle formulas**: - \( \sin 2\theta = 2 \sin \theta \cos \theta \) - \( \cos 2\theta = 2 \cos^2 \theta - 1 \) From \( \cos 2\theta = -\frac{1}{\sqrt{6}} \): \[ 2 \cos^2 \theta - 1 = -\frac{1}{\sqrt{6}} \implies 2 \cos^2 \theta = 1 - \frac{1}{\sqrt{6}} \implies \cos^2 \theta = \frac{1}{2} \left(1 - \frac{1}{\sqrt{6}}\right) \] ### Step 4: Calculate \( \sin \theta \cos \theta \) Using \( \sin 2\theta = 2 \sin \theta \cos \theta \): \[ 2 \sin \theta \cos \theta = \frac{\sqrt{30}}{6} \implies \sin \theta \cos \theta = \frac{\sqrt{30}}{12} \] ### Summary of Results 1. \( \sin 2\theta = \frac{\sqrt{30}}{6} \) 2. \( \sin \theta \cos \theta = \frac{\sqrt{30}}{12} \) 3. \( 2 \cos^2 \theta = 1 - \frac{1}{\sqrt{6}} \) Thus, the values are: \[ \begin{array}{lll} \sin 2\theta = \frac{\sqrt{30}}{6} & \sin \theta \cos \theta = \frac{\sqrt{30}}{12} & 2 \cos^2 \theta = 1 - \frac{1}{\sqrt{6}} \\ \end{array} \]