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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $

Vedclass · Accepted Answer

$x>y$

Vedclass · Answer

(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

Similar Questions

The sum $\sum\limits_{n = 1}^\infty {{\cot }^{ - 1}} \left( {\frac{{2\left( {\sum\limits_{k = 1}^n k } \right) - 1}}{3}} \right)$ is equal to

The imaginary part of $\tan^{-1}\left(\frac{5i}{3}\right)$ is

If $u = \tan^{-1} \left( \frac{\sqrt{1 + x^2} - 1}{x} \right)$ and $v = 2 \tan^{-1} x$,then $\frac{du}{dv}$ is equal to

If $p, q, r$ are in $G.P.$ and $\tan^{-1} p, \tan^{-1} q, \tan^{-1} r$ are in $A.P.$,then $p, q, r$ satisfy the relation:

The number of solutions of the equation $\sin \left[2 \cos^{-1} \left\{\cot \left(2 \tan^{-1} x\right)\right\}\right] = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module