Solve : `2pi-(sin^(-1)(4/5)+sin^(-1)(5/13)+sin^(-1)(16/65))=`

Answer

To solve the equation \( 2\pi - \left( \sin^{-1}\left(\frac{4}{5}\right) + \sin^{-1}\left(\frac{5}{13}\right) + \sin^{-1}\left(\frac{16}{65}\right) \right) \), we will follow these steps:

### Step 1: Use the formula for the sum of inverse sine functions
We can use the formula for the sum of two inverse sine functions:
\[ \sin^{-1}(x) + \sin^{-1}(y) = \sin^{-1}\left(x \sqrt{1 - y^2} + y \sqrt{1 - x^2}\right) \]
This formula is valid when \( x^2 + y^2 \leq 1 \).

### Step 2: Calculate \( \sin^{-1}\left(\frac{4}{5}\right) + \sin^{-1}\left(\frac{5}{13}\right) \)
Let \( x = \frac{4}{5} \) and \( y = \frac{5}{13} \).

1. Calculate \( x^2 + y^2 \):
\[ x^2 = \left(\frac{4}{5}\right)^2 = \frac{16}{25}, \quad y^2 = \left(\frac{5}{13}\right)^2 = \frac{25}{169} \]
\[ x^2 + y^2 = \frac{16}{25} + \frac{25}{169} \]
To add these fractions, find a common denominator:
\[ \text{LCM of } 25 \text{ and } 169 = 4225 \]
\[ x^2 + y^2 = \frac{16 \cdot 169}{4225} + \frac{25 \cdot 25}{4225} = \frac{2704 + 625}{4225} = \frac{3329}{4225} < 1 \]
Thus, we can use the formula.

2. Calculate \( \sqrt{1 - y^2} \) and \( \sqrt{1 - x^2} \):
\[ \sqrt{1 - y^2} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13} \]
\[ \sqrt{1 - x^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

3. Substitute into the formula:
\[ \sin^{-1}\left(\frac{4}{5}\right) + \sin^{-1}\left(\frac{5}{13}\right) = \sin^{-1}\left(\frac{4}{5} \cdot \frac{12}{13} + \frac{5}{13} \cdot \frac{3}{5}\right) \]
Calculate:
\[ \frac{4 \cdot 12}{65} + \frac{5 \cdot 3}{65} = \frac{48 + 15}{65} = \frac{63}{65} \]
Thus,
\[ \sin^{-1}\left(\frac{4}{5}\right) + \sin^{-1}\left(\frac{5}{13}\right) = \sin^{-1}\left(\frac{63}{65}\right) \]

### Step 3: Add the third term
Now, we need to add \( \sin^{-1}\left(\frac{63}{65}\right) + \sin^{-1}\left(\frac{16}{65}\right) \).

Let \( z = \frac{63}{65} \) and \( w = \frac{16}{65} \).

1. Calculate \( z^2 + w^2 \):
\[ z^2 = \left(\frac{63}{65}\right)^2 = \frac{3969}{4225}, \quad w^2 = \left(\frac{16}{65}\right)^2 = \frac{256}{4225} \]
\[ z^2 + w^2 = \frac{3969 + 256}{4225} = \frac{4225}{4225} = 1 \]
Since \( z^2 + w^2 = 1 \), we can use the identity:
\[ \sin^{-1}(z) + \sin^{-1}(w) = \frac{\pi}{2} \]

### Step 4: Final calculation
Now we have:
\[ \sin^{-1}\left(\frac{4}{5}\right) + \sin^{-1}\left(\frac{5}{13}\right) + \sin^{-1}\left(\frac{16}{65}\right) = \frac{\pi}{2} \]
Thus,
\[ 2\pi - \left(\frac{\pi}{2}\right) = 2\pi - \frac{\pi}{2} = \frac{4\pi}{2} - \frac{\pi}{2} = \frac{3\pi}{2} \]

### Final Answer
\[ \boxed{\frac{3\pi}{2}} \]