(i) sin^{-1} \((\frac{1}{2})\) = \(\frac{π}{6}\)

(ii) Let sin^{-1} (\(\frac{3}{5}\)) = x and sin^{-1} (\(\frac{8}{17}\)) = y

sin x = \(\frac{3}{5}\) ⇒ cos x = \(\sqrt{1-sin^2x}\)

= \(\sqrt{1- (\frac {3}{5})^2}\) = \(\frac{4}{5}\) ; sin y = \(\frac{8}{17}\)

cos (x - y) = cos x cos y + sin x sin y

=