Solve the following
` sin^(-1) . 1/5 + sin^(-1) . 2/3 = sin^(-1) x `

Answer

To solve the equation \( \sin^{-1} \left( \frac{1}{5} \right) + \sin^{-1} \left( \frac{2}{3} \right) = \sin^{-1} x \), we will use the property of inverse trigonometric functions. The property states that:

\[ \sin^{-1} a + \sin^{-1} b = \sin^{-1} \left( a \sqrt{1 - b^2} + b \sqrt{1 - a^2} \right) \]

### Step 1: Identify \( a \) and \( b \)
Let \( a = \frac{1}{5} \) and \( b = \frac{2}{3} \).

### Step 2: Calculate \( \sqrt{1 - b^2} \)
First, we need to calculate \( \sqrt{1 - b^2} \):

\[ b^2 = \left( \frac{2}{3} \right)^2 = \frac{4}{9} \]
\[ 1 - b^2 = 1 - \frac{4}{9} = \frac{5}{9} \]
\[ \sqrt{1 - b^2} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \]

### Step 3: Calculate \( \sqrt{1 - a^2} \)
Next, we calculate \( \sqrt{1 - a^2} \):

\[ a^2 = \left( \frac{1}{5} \right)^2 = \frac{1}{25} \]
\[ 1 - a^2 = 1 - \frac{1}{25} = \frac{24}{25} \]
\[ \sqrt{1 - a^2} = \sqrt{\frac{24}{25}} = \frac{\sqrt{24}}{5} \]

### Step 4: Substitute into the property
Now we substitute \( a \), \( b \), \( \sqrt{1 - b^2} \), and \( \sqrt{1 - a^2} \) into the property:

\[ \sin^{-1} \left( \frac{1}{5} \right) + \sin^{-1} \left( \frac{2}{3} \right) = \sin^{-1} \left( \frac{1}{5} \cdot \frac{\sqrt{5}}{3} + \frac{2}{3} \cdot \frac{\sqrt{24}}{5} \right) \]

### Step 5: Simplify the expression
Now we simplify the right-hand side:

\[ = \sin^{-1} \left( \frac{\sqrt{5}}{15} + \frac{2\sqrt{24}}{15} \right) \]

Combine the terms:

\[ = \sin^{-1} \left( \frac{\sqrt{5} + 2\sqrt{24}}{15} \right) \]

### Step 6: Final expression for \( x \)
Thus, we have:

\[ x = \frac{\sqrt{5} + 2\sqrt{24}}{15} \]

### Final Answer
The solution to the equation is:

\[ x = \frac{\sqrt{5} + 4\sqrt{6}}{15} \]

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