1. The problem asks to find the sine and cosine of the angle $\alpha$ given the point $(2,4)$ on its terminal side. 2. Recall that for a point $(x,y)$ on the terminal side of an angle $\alpha$ in standard position, the sine and cosine are given by: $$\sin \alpha = \frac{y}{r} \quad \text{and} \quad \cos \alpha = \frac{x}{r}$$ where $r = \sqrt{x^2 + y^2}$ is the distance from the origin to the point. 3. Calculate $r$: $$r = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$$ 4. Calculate $\sin \alpha$: $$\sin \alpha = \frac{4}{2\sqrt{5}} = \frac{\cancel{4}}{\cancel{2}\sqrt{5}} = \frac{2}{\sqrt{5}}$$ 5. Calculate $\cos \alpha$: $$\cos \alpha = \frac{2}{2\sqrt{5}} = \frac{\cancel{2}}{\cancel{2}\sqrt{5}} = \frac{1}{\sqrt{5}}$$ 6. To rationalize the denominators: $$\sin \alpha = \frac{2}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$$ $$\cos \alpha = \frac{1}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{5}}{5}$$ Final answer: $$\sin \alpha = \frac{2\sqrt{5}}{5}, \quad \cos \alpha = \frac{\sqrt{5}}{5}$$