Evaluate ` sin^(-1) . 4/5 + sin^(-1) . 5/13 + sin^(-1) . 16/65`

Answer

To evaluate the expression \( \sin^{-1} \left( \frac{4}{5} \right) + \sin^{-1} \left( \frac{5}{13} \right) + \sin^{-1} \left( \frac{16}{65} \right) \), we will use the formula for the sum of inverse sine functions.

### Step-by-Step Solution:

1. **Identify the formula**:
The formula for the sum of two inverse sine functions is:
\[ \sin^{-1}(x) + \sin^{-1}(y) = \sin^{-1} \left( x \sqrt{1 - y^2} + y \sqrt{1 - x^2} \right) \]
provided \( x, y \) are in the appropriate range.

2. **Calculate \( \sin^{-1} \left( \frac{4}{5} \right) + \sin^{-1} \left( \frac{5}{13} \right) \)**:
Let \( x = \frac{4}{5} \) and \( y = \frac{5}{13} \).
- Calculate \( \sqrt{1 - y^2} \):
\[ y^2 = \left( \frac{5}{13} \right)^2 = \frac{25}{169} \quad \Rightarrow \quad 1 - y^2 = 1 - \frac{25}{169} = \frac{144}{169} \quad \Rightarrow \quad \sqrt{1 - y^2} = \frac{12}{13} \]
- Calculate \( \sqrt{1 - x^2} \):
\[ x^2 = \left( \frac{4}{5} \right)^2 = \frac{16}{25} \quad \Rightarrow \quad 1 - x^2 = 1 - \frac{16}{25} = \frac{9}{25} \quad \Rightarrow \quad \sqrt{1 - x^2} = \frac{3}{5} \]
- Substitute into the formula:
\[ \sin^{-1} \left( \frac{4}{5} \right) + \sin^{-1} \left( \frac{5}{13} \right) = \sin^{-1} \left( \frac{4}{5} \cdot \frac{12}{13} + \frac{5}{13} \cdot \frac{3}{5} \right) \]
\[ = \sin^{-1} \left( \frac{48}{65} + \frac{15}{65} \right) = \sin^{-1} \left( \frac{63}{65} \right) \]

3. **Now add \( \sin^{-1} \left( \frac{63}{65} \right) + \sin^{-1} \left( \frac{16}{65} \right) \)**:
Let \( z = \frac{16}{65} \).
- Calculate \( \sqrt{1 - z^2} \):
\[ z^2 = \left( \frac{16}{65} \right)^2 = \frac{256}{4225} \quad \Rightarrow \quad 1 - z^2 = 1 - \frac{256}{4225} = \frac{3969}{4225} \quad \Rightarrow \quad \sqrt{1 - z^2} = \frac{63}{65} \]
- Substitute into the formula:
\[ \sin^{-1} \left( \frac{63}{65} \right) + \sin^{-1} \left( \frac{16}{65} \right) = \sin^{-1} \left( \frac{63}{65} \cdot \frac{63}{65} + \frac{16}{65} \cdot \frac{3}{5} \right) \]
\[ = \sin^{-1} \left( \frac{3969}{4225} + \frac{48}{65} \right) \]
\[ = \sin^{-1} \left( \frac{3969 + 48 \cdot 65}{4225} \right) \]
\[ = \sin^{-1} \left( \frac{3969 + 3120}{4225} \right) = \sin^{-1} \left( \frac{7089}{4225} \right) \]

4. **Final Result**:
Since \( \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \), we can conclude:
\[ \sin^{-1} \left( \frac{63}{65} \right) + \sin^{-1} \left( \frac{16}{65} \right) = \frac{\pi}{2} \]

### Final Answer:
\[ \sin^{-1} \left( \frac{4}{5} \right) + \sin^{-1} \left( \frac{5}{13} \right) + \sin^{-1} \left( \frac{16}{65} \right) = \frac{\pi}{2} \]