Question

The maximum value of sin² (120° + θ) + sin² (120° − θ) is

(a) 1/2
(b) 3/2
(c) 1/4
(d) 3/4

Answer 1

(b) $\frac{3}{2}$

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