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.)

Answer

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Answer

Thus, the exact value of $\sin \alpha$ is $\boxed{0.316227766016838}$.

Steps

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 $\boxed{0.316227766016838}$.