Problem

The key is using double-angle formulas, which let you express $\sin 2A$ and $\cos 2A$ directly in terms of $\sin A$ and $\cos A$. First, find $\cos A$ using the Pythagorean identity—since $A$ is acute, $\cos A$ is positive. Then apply the formulas $\sin 2A = 2\sin A \cos A$, $\cos 2A = \cos^2 A - \sin^2 A$, and $\tan 2A = \frac{\sin 2A}{\cos 2A}$. For $\sin 4A$, treat it as $\sin(2 \cdot 2A)$ and reuse the double-angle formula with your results.

The key is using double-angle formulas, which let you express $\sin 2A$ and $\cos 2A$ directly in terms of $\sin A$ and $\cos A$. First, find $\cos A$ using the Pythagorean identity—since $A$ is acute, $\cos A$ is positive. Then apply the formulas $\sin 2A = 2\sin A \cos A$, $\cos 2A = \cos^2 A - \sin^2 A$, and $\tan 2A = \frac{\sin 2A}{\cos 2A}$. For $\sin 4A$, treat it as $\sin(2 \cdot 2A)$ and reuse the double-angle formula with your results.