Rewrite $\sin \left(2 \cos ^{-1} w\right)$ as an algebraic expression in $w$.

Step 1 ：We know that \(\sin(\cos^{-1} w)\) and \(\cos(\cos^{-1} w)\) can be expressed in terms of \(w\). Specifically, \(\cos(\cos^{-1} w) = w\) and \(\sin(\cos^{-1} w) = \sqrt{1 - w^2}\), because \(\sin^2(\theta) + \cos^2(\theta) = 1\) for any angle \(\theta\).

Step 2 ：Substituting these expressions into the formula gives us \(2\sin(\cos ^{-1} w)\cos(\cos ^{-1} w) = 2\sqrt{1 - w^2} \cdot w\).

Step 3 ：Final Answer: \(\sin \left(2 \cos ^{-1} w\right) = \boxed{2w\sqrt{1 - w^2}}\)