Question

If ${\cos }^2({60}^0+\times )-{\cos }^2({60}^0-\times )\in [-\frac{m}{2},\frac{m}{2}]$,
then $m=$

• A. $1$
• B. $3$
• C. $-\sqrt{3}$
• D. $\sqrt{3}$

Answer 1

Answer: (D)

$\sqrt{3}$

Let $f\left(x\right)={\cos }^2(60+x)-{\cos }^2({60}^o-x)$
$=\left[\cos \right(60+x)+\cos (60-x\left)\right]\left[\cos \right(60+x)-\cos (60-x\left)\right]$
$=\left(2\cos 60\cos x\right)(-2\sin 60\sin x)$

$f\left(x\right)=-\frac{\sqrt{3}}{2}\sin 2x$
Range of $f\left(x\right)ϵ[-\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2}]=[-\frac{m}{2},\frac{m}{2}]$
$m=\sqrt{3}$
$[\because \cos u+\cos v=2\cos \left(\frac{u+v}{2}\right)\cos \left(\frac{u-v}{2}\right),\cos u-\cos v=-2\sin \left(\frac{u+v}{2}\right)\sin \left(\frac{u-v}{2}\right)]$