Prove that $\sin^{-1}\left(\dfrac{4}{5}\right)+\sin^{-1}\left(\dfrac{5}{13}\right)+\sin^{-1}\left(\dfrac{16}{65}\right)=\dfrac{\pi}{2}$.

Proved: the sum equals $\dfrac{\pi}{2}$.

CBSE Class 12 · 2009 · Inverse Trigonometric FunctionsLong Answer

Prove that

sin^{- 1} (\frac{4}{5}) + sin^{- 1} (\frac{5}{13}) + sin^{- 1} (\frac{16}{65}) = \frac{π}{2}

Answer: Proved: the sum equals $\frac{π}{2}$ .

Combine the first two using

sin^{- 1} a + sin^{- 1} b = sin^{- 1} (a 1 - b^{2} + b 1 - a^{2})

, with

a = \frac{4}{5}, b = \frac{5}{13}

1 - \frac{25}{169} = \frac{12}{13}, 1 - \frac{16}{25} = \frac{3}{5}

sin^{- 1} \frac{4}{5} + sin^{- 1} \frac{5}{13} = sin^{- 1} (\frac{4}{5} \cdot \frac{12}{13} + \frac{5}{13} \cdot \frac{3}{5}) = sin^{- 1} (\frac{48 + 15}{65}) = sin^{- 1} \frac{63}{65}

sin ϕ = \frac{63}{65}

then

cos ϕ = \frac{16}{65}

, so

sin^{- 1} \frac{63}{65} = cos^{- 1} \frac{16}{65}

. Hence

sin^{- 1} \frac{63}{65} + sin^{- 1} \frac{16}{65} = cos^{- 1} \frac{16}{65} + sin^{- 1} \frac{16}{65} = \frac{π}{2}

\frac{π}{2}

CBSE Class 12 Board · 2009 · Subjective