Question

The minimum and maximum values of ${\cos }^2({60}^0-x)-{\cos }^2({60}^0+x)$ are

• A. $-\frac{1}{2},-\frac{3}{2}$
• B. $\frac{1}{2},\frac{3}{2}$
• C. $-\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2}$
• D. $\frac{1}{2},1$

Answer 1

The correct option is C $-\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2}$

$f\left(x\right)=co{s}^2({60}^o-x)-co{s}^2({60}^o+x)$
$=\left(cos\right(60-x)-cos(60+x\left)\right)\left(cos\right(60+x)+cos(60-x\left)\right)$
$=[-2sin60sin(-x\left)\right]\left[2cos60cosx\right]$
$=2\times \frac{\sqrt{3}}{2}sinxcosx$
$f\left(x\right)=\frac{\sqrt{3}}{2}sin2x$
range of $f\left(x\right)ϵ[-\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2}]$