AMC 12 Mock Exam 4 Problem 14

Since $\theta$ is acute with $\tan\theta=2$, a right triangle with legs $1$ and $2$ gives $\sin\theta=\frac{2}{\sqrt5}$. Then $$\sin3\theta=3\sin\theta-4\sin^3\theta=\frac{6}{\sqrt5}-\frac{32}{5\sqrt5}=\frac{30-32}{5\sqrt5}=-\frac{2}{5\sqrt5}=\boxed{-\tfrac{2\sqrt5}{25}}.$$ (The negative sign makes sense: $\theta>45^\circ$ forces $3\theta>180^\circ$, and $\theta<90^\circ$ gives $3\theta<270^\circ$.)