Given that $\cos 2 \alpha=\frac{4}{5}$ and $\alpha$ terminates in quadrant I, find the exact value of $\sin \alpha$ \[ \sin \alpha= \] (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.)

Step 1 ：We are given that \(\cos 2 \alpha=\frac{4}{5}\) and \(\alpha\) is in quadrant I.

Step 2 ：We can use the double angle formula for cosine, which is \(\cos 2 \alpha = 1 - 2 \sin^2 \alpha\).

Step 3 ：Setting this equal to \(\frac{4}{5}\), we get the equation \(1 - 2 \sin^2 \alpha = \frac{4}{5}\).

Step 4 ：Solving this equation for \(\sin \alpha\), we get two solutions: \(-0.316227766016838\) and \(0.316227766016838\).

Step 5 ：Since \(\alpha\) is in quadrant I, where sine is positive, we discard the negative solution.

Step 6 ：Thus, the exact value of \(\sin \alpha\) is \(\boxed{0.316227766016838}\).