Let ${{{{\sin }^{ - 1}}x} \over \alpha } = {{{{\cos }^{ - 1}}x} \over \beta } = k \Rightarrow {\sin ^{ - 1}}x + {\cos ^{ - 1}}x = k(\alpha + \beta )$ $\Rightarrow \alpha + \beta = {\pi \over {2k}}$ Now, ${{2\pi \,\alpha } \over {\alpha + \beta }} = {{2\pi \,\alpha } \over {{\pi \over {2k}}}} = 4k\alpha = 4{\sin ^{ - 1}}x$ Here $\sin \left( {{{2\pi \,\alpha } \over {\alpha + \beta }}} \right) = \sin (4{\sin ^{ - 1}}x)$ Let ${\sin ^{ - 1}}x = \theta$ $\because$ $x \in \left( {0,{1 \over {\sqrt 2 }}} \right) \Rightarrow \theta \in \left( {0,{\pi \over 4}} \right)$ $\Rightarrow x = \sin \theta$ $\Rightarrow \cos \theta = \sqrt {1 - {x^2}}$ $\Rightarrow \sin 2\theta = 2x\,.\,\sqrt {1 - {x^2}}$ $\Rightarrow \cos 2\theta = \sqrt {1 - 4{x^2}(1 - {x^2})} = \sqrt {{{(2{x^2} - 1)}^2}} = 1 - 2{x^2}$ $\because$ $\left( {\cos 2\theta > 0\,\mathrm{as}\,2\theta \in \left( {0,{\pi \over 2}} \right)} \right)$ $\Rightarrow \sin 4\theta = 2\,.\,2x\sqrt {1 - {x^2}} (1 - 2{x^2})$ $= 4x\sqrt {1 - {x^2}} (1 - 2{x^2})$