Question

if $2{\sin }^2\left(\left(\frac{\pi }{2}\right){\cos }^{2}x\right)=1-\cos \left(\pi {\sin }^{2}2x\right),x\ne (2n+1)\frac{\pi }{2},nϵI$,then $\cos 2x$ is equal to

• A. $\frac{1}{5}$
• B. $\frac{1}{2}$
• C. $\frac{4}{5}$
• D. $1$

Answer 1

The correct option is B $\frac{1}{2}$

$2{\sin }^2\left(\frac{\pi }{2}{\cos }^{2}x\right)=1-\cos \left(\pi {\sin }^{2}2x\right),x\ne (2n+1)\frac{\pi }{2},n\in I$
$\Rightarrow 2{\sin }^2\left(\frac{\pi }{2}{\cos }^{2}x\right)=2{\sin }^2\left(\frac{\pi }{2}{\sin }^{2}2x\right)\Rightarrow \frac{\pi }{2}{\cos }^{2}x=n\pi \pm \frac{\pi }{2}{\sin }^{2}2x\Rightarrow {\cos }^{2}x=2n\pm {\sin }^{2}2x$
By observation, only $n=0$ holds and
${\cos }^{2}x={\sin }^{2}2x$
$\to {\cos }^{2}x=0$ or ${\sin }^{2}x=\frac{1}{4}$
but, given $x\ne (2n+1)\frac{\pi }{2}$
$\therefore \cos 2x=1-2{\sin }^{2}x=\frac{1}{2}$