Question

Let $x=\sin \left(2{\tan }^{-1}\alpha \right)$ and $y=\sin \left(\frac{1}{2}{\tan }^{-1}\frac{4}{3}\right)$. If $S=\left\{\alpha \in R:y^2=1-x\right\}$, then $\sum _{\alpha \in S}16{\alpha }^3$ is equal to ______

Answer 1

Answer: 130

$x=\sin \left(2{\tan }^{-1}\alpha \right)=\frac{2\tan \theta }{1+{\tan }^2\theta }=\frac{2\alpha }{1+{\alpha }^2}$
${\tan }^{-1}\alpha =\theta \Rightarrow \tan \theta =\alpha$
$y^2={\sin }^2\left(\frac{1}{2}{\tan }^{-1}\frac{4}{3}\right)=\frac{1}{5}$
$y^2+x=1\Rightarrow \frac{1}{5}+\frac{2\alpha }{1+{\alpha }^2}=1$
$\frac{2\alpha }{1+{\alpha }^2}=\frac{4}{5}$
$(2\alpha -1)(\alpha -2)=0$
$\Rightarrow 2{\alpha }^2-5\alpha +2=0$
$\therefore \alpha =2\text{ or }\frac{1}{2}$
$S=\left\{2,\frac{1}{2}\right\}$
$\sum _{\alpha \in S}16{\alpha }^3=16\left(8+\frac{1}{8}\right)=130$