If $0 < x < {1 \over {\sqrt 2 }}$ and ${{{{\sin }^{ - 1}}x} \over \alpha } = {{{{\cos }^{ - 1}}x} \over \beta }$, then the value of $\sin \left( {{{2\pi \alpha } \over {\alpha + \beta }}} \right)$ is :

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