If `"sin"^(-1)(1)/(3)+"sin"^(-1)(2)/(3)=sin^(-1)x`, then the value of x is

Options

A. 0
B. `((sqrt(5)-4sqrt(2)))/9`
C. `((sqrt(5)+4sqrt(2)))/9`
D. `(pi)/2`

Answer

To solve the equation \( \sin^{-1}\left(\frac{1}{3}\right) + \sin^{-1}\left(\frac{2}{3}\right) = \sin^{-1}(x) \), we will use the identity for the sum of two inverse sine functions.

### Step 1: Identify the values of A and B
Let:
- \( A = \frac{1}{3} \)
- \( B = \frac{2}{3} \)

### Step 2: Use the identity for \( \sin^{-1}(A) + \sin^{-1}(B) \)
The identity states:
\[ \sin^{-1}(A) + \sin^{-1}(B) = \sin^{-1}\left(A \sqrt{1 - B^2} + B \sqrt{1 - A^2}\right) \]

### Step 3: Calculate \( \sqrt{1 - B^2} \) and \( \sqrt{1 - A^2} \)
First, we calculate \( \sqrt{1 - B^2} \):
\[ B^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \implies 1 - B^2 = 1 - \frac{4}{9} = \frac{5}{9} \implies \sqrt{1 - B^2} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \]

Now, we calculate \( \sqrt{1 - A^2} \):
\[ A^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9} \implies 1 - A^2 = 1 - \frac{1}{9} = \frac{8}{9} \implies \sqrt{1 - A^2} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3} \]

### Step 4: Substitute into the identity
Now substitute \( A \), \( B \), \( \sqrt{1 - B^2} \), and \( \sqrt{1 - A^2} \) into the identity:
\[ \sin^{-1}\left(\frac{1}{3} \cdot \frac{\sqrt{5}}{3} + \frac{2}{3} \cdot \frac{2\sqrt{2}}{3}\right) \]

### Step 5: Simplify the expression
Calculate each term:
1. First term:
\[ \frac{1}{3} \cdot \frac{\sqrt{5}}{3} = \frac{\sqrt{5}}{9} \]
2. Second term:
\[ \frac{2}{3} \cdot \frac{2\sqrt{2}}{3} = \frac{4\sqrt{2}}{9} \]

Combine these:
\[ \sin^{-1}\left(\frac{\sqrt{5}}{9} + \frac{4\sqrt{2}}{9}\right) = \sin^{-1}\left(\frac{\sqrt{5} + 4\sqrt{2}}{9}\right) \]

### Step 6: Set equal to \( \sin^{-1}(x) \)
Since \( \sin^{-1}\left(\frac{\sqrt{5} + 4\sqrt{2}}{9}\right) = \sin^{-1}(x) \), we have:
\[ x = \frac{\sqrt{5} + 4\sqrt{2}}{9} \]

### Conclusion
The value of \( x \) is:
\[ x = \frac{\sqrt{5} + 4\sqrt{2}}{9} \]