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

Correct Answer: 130

To solve the problem, we start by expressing $x=\sin\left(2\tan^{-1}\alpha\right)$. Use the identity $\sin(2\theta)=\frac{2\tan\theta}{1+\tan^2\theta}$. Thus, $x=\frac{2\alpha}{1+\alpha^2}$.
Next, consider $y=\sin\left(\frac{1}{2}\tan^{-1}\frac{4}{3}\right)$. Note the identity $\sin\left(\frac{\theta}{2}\right)=\sqrt{\frac{1-\cos\theta}{2}}$ for $\theta=\tan^{-1}\frac{4}{3}$. Determine $\cos(\tan^{-1}\frac{4}{3})=\frac{1}{\sqrt{1+\left(\frac{4}{3}\right)^2}}=\frac{3}{5}$.
Next, $y=\sqrt{\frac{1-\frac{3}{5}}{2}}=\sqrt{\frac{1}{5}}$. Therefore, $y^2=\frac{1}{5}$.
Set $y^2=1-x$, giving $1-x=\frac{1}{5}\Rightarrow x=\frac{4}{5}$. Solve $\frac{2\alpha}{1+\alpha^2}=\frac{4}{5}$, leading to $10\alpha=4(1+\alpha^2)$. Rearrange to $4\alpha^2-10\alpha+4=0$. Solve via quadratic formula: $\alpha=\frac{10\pm\sqrt{100-64}}{8}=\frac{10\pm6}{8}$. Thus, $\alpha=\frac{4}{8}=\frac{1}{2}$ and $\alpha=\frac{16}{8}=2$.
We compute $\displaystyle\sum_{\alpha\in S}16\alpha^3=16\left[\left(\frac{1}{2}\right)^3+2^3\right]=16\left[\frac{1}{8}+8\right]=16\left[\frac{1}{8}+\frac{64}{8}\right]=16\times\frac{65}{8}=130$.
Hence, the sum is 130, which falls within the provided range of 130,130.