c) If \( 2 \tan 2 \theta=-\sqrt{5} \) with \( 0^{\circ}<2 \theta<180^{\circ} \), determine the value of the following without th of a calculator and with the aid of a diagram: \( \begin{array}{ll}\text { (1) } \sin 2 \theta & \text { (2) } \sin \theta \cos \theta \\ \text { (4) } \cos \theta & \end{array} \)

To solve the problem, we start with the equation given: \[ 2 \tan 2\theta = -\sqrt{5} \] From this, we can express \(\tan 2\theta\): \[ \tan 2\theta = -\frac{\sqrt{5}}{2} \] ### Step 1: Determine the angle \(2\theta\) Since \(0^{\circ} < 2\theta < 180^{\circ}\), we know that \(2\theta\) is in the first or second quadrant. The tangent function is negative in the second quadrant. Therefore, we can find \(2\theta\) as follows: Let \(2\theta = 180^{\circ} - \alpha\), where \(\tan \alpha = \frac{\sqrt{5}}{2}\). To find \(\alpha\), we can use the tangent inverse: \[ \alpha = \tan^{-1}\left(\frac{\sqrt{5}}{2}\right) \] Thus, \[ 2\theta = 180^{\circ} - \alpha \] ### Step 2: Calculate \(\sin 2\theta\) Using the identity for sine in terms of tangent: \[ \sin 2\theta = \frac{2 \tan 2\theta}{1 + \tan^2 2\theta} \] First, we calculate \(\tan^2 2\theta\): \[ \tan^2 2\theta = \left(-\frac{\sqrt{5}}{2}\right)^2 = \frac{5}{4} \] Now substituting into the sine formula: \[ \sin 2\theta = \frac{2 \left(-\frac{\sqrt{5}}{2}\right)}{1 + \frac{5}{4}} = \frac{-\sqrt{5}}{\frac{9}{4}} = -\frac{4\sqrt{5}}{9} \] ### Step 3: Calculate \(\sin \theta \cos \theta\) Using the double angle identity: \[ \sin 2\theta = 2 \sin \theta \cos \theta \] Thus, \[ \sin \theta \cos \theta = \frac{\sin 2\theta}{2} = \frac{-\frac{4\sqrt{5}}{9}}{2} = -\frac{2\sqrt{5}}{9} \] ### Step 4: Calculate \(\cos \theta\) Using the identity: \[ \cos^2 2\theta + \sin^2 2\theta = 1 \] We already have \(\sin 2\theta\), so we can find \(\cos 2\theta\): \[ \cos 2\theta = \sqrt{1 - \sin^2 2\theta} = \sqrt{1 - \left(-\frac{4\sqrt{5}}{9}\right)^2} \] Calculating \(\sin^2 2\theta\): \[ \sin^2 2\theta = \left(-\frac{4\sqrt{5}}{9}\right)^2 = \frac{80}{81} \] Now substituting back: \[ \cos 2\theta = \sqrt{1 - \frac{80}{81}} = \sqrt{\frac{1}{81}} = \frac{1}{9} \] Using the double angle formula for cosine: \[ \cos 2\theta = 2 \cos^2 \theta - 1 \] Setting this equal to \(\frac{1}{9}\): \[ 2 \cos^2 \theta - 1 = \frac{1}{9} \] Solving for \(\cos^2 \theta\): \[ 2 \cos^2 \theta = \frac{1}{9} + 1 = \frac{10}{9} \] \[ \cos^2 \theta = \frac{5}{9} \] Thus, \[ \cos \theta = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \] ### Summary of Results 1. \(\sin 2\theta = -\frac{4\sqrt{5}}{9}\) 2. \(\sin \theta \cos \theta = -\frac{2\sqrt{5}}{9}\) 3. \(\cos \theta = \frac{\sqrt{5}}{3}\) Final answers: \[ \begin{array}{ll} (1) & \sin 2\theta = -\frac{4\sqrt{5}}{9} \\ (2) & \sin \theta \cos \theta = -\frac{2\sqrt{5}}{9} \\ (4) & \cos \theta = \frac{\sqrt{5}}{3} \\ \end{array} \]