Drag and drop the correct answer into the space(s) provided. If \( \tan \theta=\frac{1}{2} \) on the interval \( \left(\pi, \frac{3 \pi}{2}\right) \), find \( \sin 2 \theta, \cos 2 \theta \), and \( \tan 2 \theta \). \( \sin \theta=-\frac{\sqrt{5}}{5} \) and \( \cos \theta=-\frac{2 \sqrt{5}}{5} \) \( \sin 2 \theta= \) \( \cos 2 \theta= \) \( \tan 2 \theta= \) \( -\frac{\sqrt{5}}{2} \)

When we have \( \tan \theta = \frac{1}{2} \), that means for some angle \( \theta \), the sine and cosine must follow that ratio. Since \( \theta \) is in the third quadrant, both sine and cosine are negative. You already have \( \sin \theta = -\frac{\sqrt{5}}{5} \) and \( \cos \theta = -\frac{2\sqrt{5}}{5} \). To find \( \sin 2\theta \), we can use the identity: \[ \sin 2\theta = 2 \sin \theta \cos \theta \] Plugging in the values: \[ \sin 2\theta = 2 \left(-\frac{\sqrt{5}}{5}\right)\left(-\frac{2\sqrt{5}}{5}\right) = 2 \cdot \frac{2 \cdot 5}{25} = \frac{20}{25} = \frac{4}{5} \] Next, for \( \cos 2\theta \): \[ \cos 2\theta = \cos^2 \theta - \sin^2 \theta \] Calculating individually, \[ \cos^2 \theta = \left(-\frac{2\sqrt{5}}{5}\right)^2 = \frac{4 \cdot 5}{25} = \frac{20}{25} = \frac{4}{5} \] \[ \sin^2 \theta = \left(-\frac{\sqrt{5}}{5}\right)^2 = \frac{5}{25} = \frac{1}{5} \] Now plug those into the formula: \[ \cos 2\theta = \frac{4}{5} - \frac{1}{5} = \frac{3}{5} \] Lastly, for \( \tan 2\theta \): \[ \tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} \] So we calculate: \[ \tan 2\theta = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3} \] Putting it all together: \[ \sin 2 \theta = \frac{4}{5} \] \[ \cos 2 \theta = \frac{3}{5} \] \[ \tan 2 \theta = \frac{4}{3} \]