Question

If $\alpha =\sin \left(\frac{{\sin }^{-1}\frac{1}{\sqrt{3}}}{3}\right)$ $\beta =\cos \left({\cos }^{-1}\left(\frac{1}{\sqrt{5}}\right)-{\sin }^{-1}\left(\frac{2}{\sqrt{5}}\right)\right)$ then $\frac{{\beta }^2}{{\left(3\alpha -4{\alpha }^3\right)}^2}$ is equal to

Answer 1

Answer: 3

Let ${\sin }^{-1}\frac{1}{\sqrt{3}}=\theta$
$\Rightarrow \sin \theta =\frac{1}{\sqrt{3}}$
$\therefore \alpha =\sin \frac{\theta }{3}$
Hence, $\sin \left(3\cdot \frac{\theta }{3}\right)=3\sin \frac{\theta }{3}-4{\sin }^3\frac{\theta }{3}=3\alpha -4{\alpha }^3$
$\therefore 3\alpha -4{\alpha }^3=\sin \theta =\frac{1}{\sqrt{3}}$
Also, $\beta =\cos \left({\cos }^{-1}\frac{1}{\sqrt{5}}-{\sin }^{-1}\frac{2}{\sqrt{5}}\right)$
$=\cos \left({\tan }^{-1}2-{\tan }^{-1}2\right)=\cos \left(0^{\circ }\right)=1$
$\therefore \frac{{\beta }^2}{{\left(3\alpha -4{\alpha }^3\right)}^2}=\frac{1}{{\left(\frac{1}{\sqrt{3}}\right)}^2}=3.$