If x=sin left(2 tan ^{-1} 2right) and y=sin l | vedclass

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(A) Given,$x=\sin \left(2 	an ^{-1} 2ight)$ and $y=\sin \left(\frac{1}{2} 	an ^{-1} \frac{4}{3}ight)$.For $x$,let $	an ^{-1} 2 = \alpha$,so $

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$x>y$

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(A) Given,$x=\sin \left(2 	an ^{-1} 2ight)$ and $y=\sin \left(\frac{1}{2} 	an ^{-1} \frac{4}{3}ight)$.For $x$,let $	an ^{-1} 2 = \alpha$,so $	an \alpha = 2$.Then $x = \sin(2\alpha) = \frac{2 	an \alpha}{1 + 	an^2 \alpha} = \frac{2(2)}{1 + 2^2} = \frac{4}{5} = 0.8$.For $y$,let $	an ^{-1} \frac{4}{3} = \beta$,so $	an \beta = \frac{4}{3}$.Using $\cos \beta = \frac{1}{\sqrt{1 + 	an^2 \beta}} = \frac{1}{\sqrt{1 + (16/9)}} = \frac{1}{\sqrt{25/9}} = \frac{3}{5}$.Then $y = \sin(\beta/2) = \sqrt{\frac{1 - \cos \beta}{2}} = \sqrt{\frac{1 - 3/5}{2}} = \sqrt{\frac{2/5}{2}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} \approx 0.447$.Since $0.8 > 0.447$,we have $x > y$.

If $x=\sin \left(2 \tan ^{-1} 2\right)$ and $y=\sin \left(\frac{1}{2} \tan ^{-1} \frac{4}{3}\right)$,then

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