Question

Prove that \[ \sin^{-1} \frac{4}{5} + \sin^{-1} \frac{5}{13} + \sin^{-1} \frac{16}{65} = \frac{\pi}{2}. \]

Hint:

Use addition formulas for inverse trigonometric functions and simplify step-by-step with known sine and cosine values.

Answer 1

Let \[ A = \sin^{-1} \frac{4}{5}, \quad B = \sin^{-1} \frac{5}{13}, \quad C = \sin^{-1} \frac{16}{65}. \] We need to show: \[ A + B + C = \frac{\pi}{2}. \] Using the addition formula for sine: \[ \sin(A + B) = \sin A \cos B + \cos A \sin B. \] Calculate \(\sin A, \cos A, \sin B, \cos B\): \[ \sin A = \frac{4}{5}, \quad \cos A = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \frac{3}{5}. \] \[ \sin B = \frac{5}{13}, \quad \cos B = \sqrt{1 - \left(\frac{5}{13}\right)^2} = \frac{12}{13}. \] Then, \[ \sin(A + B) = \frac{4}{5} \cdot \frac{12}{13} + \frac{3}{5} \cdot \frac{5}{13} = \frac{48}{65} + \frac{15}{65} = \frac{63}{65}. \] Therefore, \[ A + B = \sin^{-1} \frac{63}{65}. \] Now, \[ (A + B) + C = \sin^{-1} \frac{63}{65} + \sin^{-1} \frac{16}{65}. \] Use sine addition formula again: \[ \sin\big((A+B) + C\big) = \sin(A+B) \cos C + \cos(A+B) \sin C. \] Calculate \(\cos C\) and \(\cos(A+B)\): \[ \cos C = \sqrt{1 - \left(\frac{16}{65}\right)^2} = \sqrt{1 - \frac{256}{4225}} = \sqrt{\frac{4225 - 256}{4225}} = \sqrt{\frac{3969}{4225}} = \frac{63}{65}. \] \[ \cos(A+B) = \sqrt{1 - \left(\frac{63}{65}\right)^2} = \sqrt{1 - \frac{3969}{4225}} = \sqrt{\frac{256}{4225}} = \frac{16}{65}. \] Then, \[ \sin\big((A+B) + C\big) = \frac{63}{65} \cdot \frac{63}{65} + \frac{16}{65} \cdot \frac{16}{65} = \frac{3969}{4225} + \frac{256}{4225} = \frac{4225}{4225} = 1. \] Thus, \[ A + B + C = \sin^{-1} 1 = \frac{\pi}{2}. \]
Final answer: \[ \boxed{ \sin^{-1} \frac{4}{5} + \sin^{-1} \frac{5}{13} + \sin^{-1} \frac{16}{65} = \frac{\pi}{2}. } \]