Trigonometry 7th Edition

$\sin \theta=\frac{2\sqrt 5}{5}$ $\cos \theta=\frac{\sqrt 5}{5}$
We can consider an arbitrary point that satisfies y=2x and lies in quadrant I. Let us consider the point (1,2). Then, $r=\sqrt {x^{2}+y^{2}}=\sqrt {1^{2}+2^{2}}=\sqrt 5$ $\sin \theta=\frac{y}{r}=\frac{2}{\sqrt 5}=\frac{2\sqrt 5}{5}$ $\cos \theta=\frac{x}{r}=\frac{1}{\sqrt 5}=\frac{\sqrt 5}{5}$