Problem
Prove: \( \sin\left(\cos^{-1}\left(\frac{3}{5}\right) + \sin^{-1}\left(\frac{5}{13}\right)\right) = \frac{63}{65} \)
Solution
Let:
\[ A = \cos^{-1}\left(\frac{3}{5}\right), \quad B = \sin^{-1}\left(\frac{5}{13}\right) \]
Step 1: Find sin A and cos A
\[ \cos A = \frac{3}{5} = \frac{\text{Base}}{\text{Hypotenuse}} \]
- Base = 3
- Hypotenuse = 5
Perpendicular:
\[ \sqrt{5^2 – 3^2} = \sqrt{25 – 9} = 4 \]
\[ \sin A = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{4}{5} \]
Step 2: Find sin B and cos B
\[ \sin B = \frac{5}{13} = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \]
- Perpendicular = 5
- Hypotenuse = 13
Base:
\[ \sqrt{13^2 – 5^2} = \sqrt{169 – 25} = 12 \]
\[ \cos B = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{12}{13} \]
Step 3: Use sin(A + B) formula
\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]
\[ = \frac{4}{5} \cdot \frac{12}{13} + \frac{3}{5} \cdot \frac{5}{13} \]
\[ = \frac{48}{65} + \frac{15}{65} = \frac{63}{65} \]
Final Result
\[ \boxed{\frac{63}{65}} \]
Explanation
We converted inverse trigonometric functions into triangle ratios and applied the sin(A+B) identity.