Let x = sin(2 tan^{-1} alpha) and y = sin(fra | vedclass

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(D) Given $x = \sin(2 	an^{-1} \alpha) = \frac{2 \alpha}{1 + \alpha^2}$.Given $y = \sin(\frac{1}{2} 	an^{-1} \frac{4}{3})$. Let $	heta = 	an^{-1}

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$130$

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(D) Given $x = \sin(2 	an^{-1} \alpha) = \frac{2 \alpha}{1 + \alpha^2}$.Given $y = \sin(\frac{1}{2} 	an^{-1} \frac{4}{3})$. Let $	heta = 	an^{-1} \frac{4}{3}$,so $	an 	heta = \frac{4}{3}$.Using the identity $\sin(\frac{	heta}{2}) = \sqrt{\frac{1 - \cos 	heta}{2}}$,where $\cos 	heta = \frac{3}{5}$,we get $y = \sqrt{\frac{1 - 3/5}{2}} = \sqrt{\frac{2/5}{2}} = \frac{1}{\sqrt{5}}$.Given $y^2 = 1 - x$,we substitute the values:$\frac{1}{5} = 1 - \frac{2 \alpha}{1 + \alpha^2}$$\frac{2 \alpha}{1 + \alpha^2} = 1 - \frac{1}{5} = \frac{4}{5}$$10 \alpha = 4(1 + \alpha^2)$$4 \alpha^2 - 10 \alpha + 4 = 0$$2 \alpha^2 - 5 \alpha + 2 = 0$$(2 \alpha - 1)(\alpha - 2) = 0$So,$\alpha = 2$ or $\alpha = \frac{1}{2}$.We need to find $\sum_{\alpha \in S} 16 \alpha^3 = 16(2^3) + 16(\frac{1}{2})^3 = 16(8) + 16(\frac{1}{8}) = 128 + 2 = 130$.

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

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