The value of `sin[2"cos"^(-1)(sqrt(5))/(3)]` is

Options

A. `(sqrt(5))/(3)`
B. `(2sqrt(5))/(3)`
C. `(4sqrt(5))/(9)`
D. `(2sqrt(5))/(9)`

Explanation

To solve the expression \( \sin\left(2 \cos^{-1}\left(\frac{\sqrt{5}}{3}\right)\right) \), we will follow these steps:

### Step 1: Use the double angle formula for sine
We know that:
\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \]
In our case, let \( \theta = \cos^{-1}\left(\frac{\sqrt{5}}{3}\right) \). Therefore:
\[ \sin\left(2 \cos^{-1}\left(\frac{\sqrt{5}}{3}\right)\right) = 2 \sin\left(\cos^{-1}\left(\frac{\sqrt{5}}{3}\right)\right) \cos\left(\cos^{-1}\left(\frac{\sqrt{5}}{3}\right)\right) \]

### Step 2: Find \( \sin(\theta) \) and \( \cos(\theta) \)
From the definition of cosine, we have:
\[ \cos(\theta) = \frac{\sqrt{5}}{3} \]
To find \( \sin(\theta) \), we can use the Pythagorean identity:
\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]
Substituting the value of \( \cos(\theta) \):
\[ \sin^2(\theta) + \left(\frac{\sqrt{5}}{3}\right)^2 = 1 \]
\[ \sin^2(\theta) + \frac{5}{9} = 1 \]
\[ \sin^2(\theta) = 1 - \frac{5}{9} = \frac{4}{9} \]
Taking the square root:
\[ \sin(\theta) = \sqrt{\frac{4}{9}} = \frac{2}{3} \]

### Step 3: Substitute back into the sine double angle formula
Now we have:
\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) = 2 \cdot \frac{2}{3} \cdot \frac{\sqrt{5}}{3} \]
Calculating this:
\[ \sin(2\theta) = \frac{4 \sqrt{5}}{9} \]

### Conclusion
Thus, the value of \( \sin\left(2 \cos^{-1}\left(\frac{\sqrt{5}}{3}\right)\right) \) is:
\[ \boxed{\frac{4\sqrt{5}}{9}} \]