Use the fundamental identities and the given information to find the exact value of $\sin \alpha$.
\[
\cos (-\alpha)=\frac{4 \sqrt{17}}{17}, \tan \alpha> 0
\]
$\sin \alpha=$
(Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.)

Step 1 ：We are given that \(\cos (-\alpha)=\frac{4 \sqrt{17}}{17}\). Since cosine is an even function, we know that \(\cos \alpha = \frac{4 \sqrt{17}}{17}\).

Step 2 ：We also know that \(\tan \alpha > 0\), which means that \(\sin \alpha\) must be positive (since \(\tan \alpha = \frac{\sin \alpha}{\cos \alpha}\)).

Step 3 ：We can use the identity \(\cos^2 \alpha + \sin^2 \alpha = 1\) to find \(\sin \alpha\). Substituting the value of \(\cos \alpha\) into the identity, we get \(\sin^2 \alpha = 1 - \cos^2 \alpha\).

Step 4 ：Solving for \(\sin \alpha\), we get two possible solutions: \(\sin \alpha = \sqrt{1 - \cos^2 \alpha}\) or \(\sin \alpha = -\sqrt{1 - \cos^2 \alpha}\).

Step 5 ：However, since we know that \(\sin \alpha\) must be positive, we discard the negative solution.

Step 6 ：Substituting the value of \(\cos \alpha\) into the equation, we get \(\sin \alpha = \sqrt{1 - \left(\frac{4 \sqrt{17}}{17}\right)^2}\).

Step 7 ：Solving this, we get \(\sin \alpha = 0.242535625036333\).

Step 8 ：Final Answer: \(\boxed{0.242535625036333}\)