Question

The maximum value of $si{n}^2({120}^{\circ }+\theta )+si{n}^2({120}^{\circ }-\theta )$ is

• A. $\frac{1}{2}$
  
• B. $\frac{3}{2}$
  
• C. $\frac{1}{4}$
  
• D. $\frac{3}{4}$

Answer 1

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


Let $f\left(\theta \right)=si{n}^2(90+30+\theta )+si{n}^2(90+30-\theta )$
$={[\frac{\sqrt{3}}{2}cos\theta -\frac{1}{2}sin\theta ]}^2+{[\frac{\sqrt{3}}{2}cos\theta +\frac{1}{2}sin\theta ]}^2$
$=\frac{3}{4}co{s}^2\theta +\frac{1}{4}si{n}^2\theta -\frac{\sqrt{3}}{2}cos\theta sin\theta +\frac{3}{4}co{s}^2\theta +\frac{1}{4}si{n}^2\theta +\frac{\sqrt{3}}{2}cos\theta sin\theta$
$=\frac{3}{2}co{s}^2\theta +\frac{1}{2}si{n}^2\theta$
$=\frac{3}{2}(1-si{n}^2\theta )+\frac{1}{2}si{n}^2\theta$
$=\frac{3}{2}-\frac{3}{2}si{n}^2\theta +\frac{1}{2}si{n}^2\theta$
$=\frac{3}{2}-si{n}^2\theta$
For $f\left(\theta \right)$ to be maximum, $si{n}^2\theta$ must have minimum value, which is 0.
$\therefore \frac{3}{2}$ is the maximum value of $f\left(\theta \right)$.