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 $\displaystyle\sum_{\alpha \in S } 16 \alpha^3$ is equal to ______

Answer 1

$ 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\} $
$\displaystyle\sum_{\alpha \in S} 16 \alpha^3=16\left(8+\frac{1}{8}\right)=130$