If \( 2 \tan 2 \theta=-\sqrt{5} \) with \( 0^{\circ}<2 \theta<180^{\circ} \), deternine the value of the following without the use of a calculator and with the aid of a diagram: \( \begin{array}{lll}\text { (1) } \sin 2 \theta & \text { (2) } \sin \theta \cos \theta & \text { (3) } 2 \cos ^{2} \theta \\ \text { (4) } \cos \theta & & \end{array} \)

To solve the problem given \( 2 \tan 2\theta = -\sqrt{5} \) and \( 0^{\circ} < 2\theta < 180^{\circ} \), we start by isolating \( \tan 2\theta \): \[ \tan 2\theta = -\frac{\sqrt{5}}{2} \] Since \( 0^{\circ} < 2\theta < 180^{\circ} \), \( 2\theta \) is in the second quadrant where the tangent function is negative. We can find \( 2\theta \) by considering the reference angle. The reference angle \( \alpha \) that corresponds to \( \tan \alpha = \frac{\sqrt{5}}{2} \) can be found using the specific values or ratios. ### Step 1: Find Sin and Cos of \( 2\theta \) Using the Pythagorean identity: \[ \tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} \] We know: \[ \sin^2 2\theta + \cos^2 2\theta = 1 \] Let's denote \( \sin 2\theta = -\sqrt{5}/k \) and \( \cos 2\theta = 2/k \) to maintain the relationship between tangent and sine/cosine: From \( 2\tan 2\theta = -\sqrt{5} \): We can write: \[ \tan 2\theta = -\frac{\sqrt{5}}{2} \] Now constructing a right triangle where the opposite side is \( \sqrt{5} \) and the adjacent side is \( -2 \): \[ r = \sqrt{(\sqrt{5})^2 + (-2)^2} = \sqrt{5 + 4} = \sqrt{9} = 3 \] Thus, \[ \sin 2\theta = \frac{\sqrt{5}}{3}, \quad \cos 2\theta = -\frac{2}{3} \] ### Step 2: Calculate Required Values 1) **Calculate \( \sin 2\theta \):** \[ \sin 2\theta = \frac{\sqrt{5}}{3} \] 2) **Calculate \( \sin \theta \cos \theta \):** Using the double angle formula for sine: \[ \sin 2\theta = 2 \sin \theta \cos \theta \] So, \[ 2 \sin \theta \cos \theta = \frac{\sqrt{5}}{3} \] Thus, \[ \sin \theta \cos \theta = \frac{\sqrt{5}}{6} \] 3) **Calculate \( 2 \cos^2 \theta \):** Using \( \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \), We have: \[ \cos 2\theta = -\frac{2}{3} \implies \cos^2 \theta = \frac{1 - \frac{2}{3}}{2} = \frac{1/3}{2} = \frac{1}{6} \] Thus, \[ 2 \cos^2 \theta = \frac{2}{6} = \frac{1}{3} \] 4) **Find \( \cos \theta \):** We use the half-angle formula for cosine: \[ \cos^2 \theta = \frac{1 + \cos 2\theta}{2} = \frac{1 - \frac{2}{3}}{2} = \frac{1/3}{2} = \frac{1}{6} \implies \cos \theta = \sqrt{\frac{1}{6}} \] Putting it all together, we get: \[ \begin{array}{lll} \text { (1) } \sin 2 \theta = \frac{\sqrt{5}}{3} \\ \text { (2) } \sin \theta \cos \theta = \frac{\sqrt{5}}{6} \\ \text { (3) } 2 \cos^2 \theta = \frac{1}{3} \\ \text { (4) } \cos \theta = \frac{1}{\sqrt{6}} \text{ or } \sqrt{\frac{1}{6}} \\ \end{array} \]