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

The correct answer is 130
x=sin(2tan−1α)=1+tan2θ2tanθ​=1+α22α​
tan−1α=θ⇒tanθ=α
y2=sin2(21​tan−134​)=51​
y2+x=1⇒51​+1+α22α​=1
1+α22α​=54​
(2α−1)(α−2)=0
⇒2α2−5α+2=0
∴α=2 or 21​
S={2,21​}
α∈S∑​16α3=16(8+81​)=130