Question

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

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

Answer 1

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

Let $f\left(x\right)=si{n}^2({60}^o-x)-si{n}^2({60}^o+x)$
$=\left[sin\right(60+x)+sin(60-x\left)\right]\left[sin\right(60-x)-sin(60+x\left)\right]$
$=\left(2sin60cosx\right)(-2cos60sinx)$
$=-4\times \frac{\sqrt{3}}{4}\times \sin x\cos x$
$f\left(x\right)=-\frac{\sqrt{3}}{2}sin2x$
max value of $f\left(x\right)=+\frac{\sqrt{3}}{2}$
min value of $f\left(x\right)=-\frac{\sqrt{3}}{2}$
formula used
$[sinu+sinv=2sin(\frac{u+v}{2}\left)cos\right(\frac{u-v}{2}),sinusinv=\frac{1}{2}\left[cos\right(u-v)-cos(u+v\left)\right],cos2x=2co{s}^{2}x-1]$