Question

If $\sin \theta \sin ({120}^0-\theta )\sin ({120}^0+\theta )\in [-\frac{m}{4}$ , $\frac{m}{4}]$, then $m=$

• A. $1$
• B. $3$
• C. $-1$
• D. $-3$

Answer 1

The correct option is A $1$

$f\left(\theta \right)=\sin \theta \sin (120-\theta )\sin (120+\theta )$
$=\sin \theta \frac{1}{2}\left[\right(\cos 2\theta -\cos \left(240\right)\left)\right]$
$=\frac{\sin \theta }{2}[2{\cos }^2\theta -\frac{1}{2}]$
$=(\frac{3}{2}-2{\sin }^2\theta )\frac{\sin \theta }{2}$
$f\left(\theta \right)=\frac{\sin 3\theta }{4}$
$\therefore$ range of $f\left(\theta \right)ϵ[-\frac{1}{4},\frac{1}{4}]=[-\frac{m}{4},\frac{m}{4}]$

$\therefore m=1$