$$ {\displaylines\surd(1-cos^2⁡(〖2cos〗^{(}-1)⁡(2/3)))\\ } $$

The expression you've provided is a bit complex, so let's break it down step by step.

We have:

\[
\sqrt{1 - \cos^2\left( 2\cos^{-1}\left( \frac{2}{3} \right) \right)}
\]

### Step-by-step process:

1. **Inside the parentheses**:  
   $\cos^{-1}\left( \frac{2}{3} \right)$ represents the inverse cosine, which gives us the angle $\theta$ such that $\cos(\theta) = \frac{2}{3}$.

2. **Double the angle**:  
   We are asked to find $2 \times \cos^{-1}\left( \frac{2}{3} \right)$. Let this angle be $\alpha = \cos^{-1}\left( \frac{2}{3} \right)$. So, $2\alpha$ will be the angle inside the cosine function.

3. **Simplify $ \cos^2(2\alpha) $**:  
   The expression now becomes $1 - \cos^2(2\alpha)$. Using the identity:
   \[
   \sin^2(\theta) = 1 - \cos^2(\theta)
   \]
   we can replace $1 - \cos^2(2\alpha)$ with $\sin^2(2\alpha)$.

4. **Final expression**:  
   The square root of $\sin^2(2\alpha)$ is simply $\sin(2\alpha)$. Therefore, the expression simplifies to:
   \[
   \sin(2 \cos^{-1}\left( \frac{2}{3} \right))
   \]

### Finding the value of $\sin(2\cos^{-1}(2/3))$:

Using the double angle identity for sine:
\[
\sin(2\alpha) = 2\sin(\alpha)\cos(\alpha)
\]
We know $\cos(\alpha) = \frac{2}{3}$. To find $\sin(\alpha)$, we use the Pythagorean identity:
\[
\sin^2(\alpha) + \cos^2(\alpha) = 1
\]
\[
\sin^2(\alpha) = 1 - \left(\frac{2}{3}\right)^2 = 1 - \frac{4}{9} = \frac{5}{9}
\]
\[
\sin(\alpha) = \frac{\sqrt{5}}{3}
\]

Now, applying the double angle formula:
\[
\sin(2\alpha) = 2 \times \frac{\sqrt{5}}{3} \times \frac{2}{3} = \frac{4\sqrt{5}}{9}
\]

So, the final value is:
\[
\boxed{\frac{4\sqrt{5}}{9}}
\]

Do you have any questions or would you like further details?

### Related questions:
1. How do we use trigonometric identities in solving such expressions?
2. What is the significance of the double-angle identity for sine?
3. How can we find the value of inverse trigonometric functions?
4. What are some common applications of $\cos^{-1}$ in trigonometry?
5. How can we simplify trigonometric expressions involving square roots?

**Tip:** Knowing the basic trigonometric identities is key to simplifying complex expressions easily!

$$ \sqrt{1-\cos^2⁡(2\cos^{-1}⁡(2/3))} $$

Learn how to simplify the expression involving trigonometric identities: $\sqrt{1 - \cos^2(2\cos^{-1}(2/3))}$. This detailed step-by-step solution covers key trigonometric concepts like inverse cosine, double-angle formulas, and the Pythagorean identity, suitable for students in Grades 10-12.

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