Question

If $f\left(x\right)=2{\sin }^2\theta +4\cos (x+\theta )\sin x\cdot \sin \theta +\cos (2x+2\theta )$ then value of $f^2\left(x\right)+f^2(\frac{\pi }{4}-x)$ is

• A. $0$
• B. $1$
• C. $-1$
• D. $x^2$

Answer 1

The correct option is B $1$

$f\left(x\right)=2{\sin }^2\theta +4\cos (x+\theta )\sin x\sin \theta +\cos (2x+2\theta )$
$=2{\sin }^2\theta +2\cos (x+\theta )\left[2\sin x\sin \theta \right]+\cos 2(x+\theta )$
$\cos (A-B)-\cos (A+B)=2\sin A\sin B$
$\therefore f\left(x\right)=2{\sin }^2\theta +2\cos (x+\theta )[\cos (x-\theta )-\cos (x+\theta )]+{\cos }^2(x+\theta )-{\sin }^2(x+\theta )$
$f\left(x\right)=2{\sin }^2\theta +2\cos (x+\theta )\cos (x-\theta )-2{\cos }^2(x+\theta )+{\cos }^2(x+\theta )-{\sin }^2(x+\theta )$
$f\left(x\right)=2{\sin }^2\theta +2\cos (x+\theta )\cos (x-\theta )-[{\cos }^2(x+\theta )+{\sin }^2(x+\theta )]$
$\cos (A+B)+\cos (A-B)=2\cos A\cos B$
$f\left(x\right)=2{\sin }^2\theta +\cos (x+\theta +x-\theta )+\cos (x+\theta -x+\theta )-1$
$f\left(x\right)=2{\sin }^2\theta +\cos 2x+\cos 2\theta -1$
$f\left(x\right)=2{\sin }^2\theta +\cos 2x-(1-\cos 2\theta )$
$f\left(x\right)=2{\sin }^2\theta +\cos 2x-2{\sin }^2\theta$
$f\left(x\right)=\cos 2x$
$f^2\left(x\right)+f^2(\frac{\pi }{4}-x)={\cos }^{2}2x+{\cos }^{2}2(\frac{\pi }{4}-x)$
$={\cos }^{2}2x+{\left[\cos (\frac{\pi }{2}-2x)\right]}^2$
$={\cos }^{2}2x+{\sin }^{2}2x$
$=1$