Rewrite $\cos \left(2 \cos ^{-1} \frac{w}{6}\right)$ as an algebraic expression in $w$.
\[
\cos \left(2 \cos ^{-1} \frac{w}{6}\right)=\square
\]

Answer

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Answer

So, $\cos \left(2 \cos ^{-1} \frac{w}{6}\right)=\boxed{2\left(\frac{w}{6}\right)^2 - 1}$

Steps

Step 1 ：Let $\theta = \cos^{-1}(\frac{w}{6})$

Step 2 ：We know that $\cos(2\theta) = 1 - 2\sin^2(\theta)$

Step 3 ：We also know that $\sin^2(\theta) = 1 - \cos^2(\theta)$

Step 4 ：So, we can substitute $\cos^2(\theta)$ with $(\frac{w}{6})^2$

Step 5 ：Substituting these values into the equation for $\cos(2\theta)$, we get $\cos(2\theta) = 1 - 2(1 - (\frac{w}{6})^2)$

Step 6 ：Simplifying this, we get $\cos(2\theta) = 2(\frac{w}{6})^2 - 1$

Step 7 ：So, $\cos \left(2 \cos ^{-1} \frac{w}{6}\right)=\boxed{2\left(\frac{w}{6}\right)^2 - 1}$