Question

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

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

Answer 1

The correct option is C $\frac{1}{2},\frac{3}{2}$

$f\left(x\right)={\sin }^2(60-x)+{\sin }^2(60+x)$

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

$[\because \sin (A+B)+\sin (A-B)=2\sin A\cos B]$ and $[2\sin A\sin B=\cos (A-B)-\cos (A+B\left)\right]$
$\Rightarrow f\left(x\right)=(2\sin 60\cos x{)}^2-[\cos 2x-\cos 120]$

$=3{\cos }^{2}x-[2{\cos }^{2}x-1+1/2]$ $[\because \sin 60=\frac{\sqrt{3}}{2}]$

$\Rightarrow f\left(x\right)={\cos }^{2}x+\frac{1}{2}$

range of $f\left(x\right)\in [\frac{1}{2},\frac{3}{2}]$