Question

The maximum value of ${\sin }^2\left(\frac{2\pi }{3}+x\right)+{\sin }^2\left(\frac{2\pi }{3}-x\right)$ is
(a) 1/2
(b) 3/2
(c) 1/4
(d) 3/4

Answer 1

(b) $\frac{3}{2}$
$\frac{2\pi }{3}=120°$
$$\begin{gathered} \mathrm{Let}f\left(x\right)={\sin }^2(90+30+x)+{\sin }^2(90+30-x) \\ ={\left[\cos (30+x)\right]}^2+{\left[\cos (30-x)\right]}^2\left[\mathrm{Using}\sin (90+A)=\cos A\right] \\ ={\left[\frac{\sqrt{3}}{2}\cos x-\frac{1}{2}\sin x\right]}^2+{\left[\frac{\sqrt{3}}{2}\cos x+\frac{1}{2}\sin x\right]}^2 \\ =\frac{3}{4}{\cos }^{2}x+\frac{1}{4}{\sin }^{2}x-\frac{\sqrt{3}}{2}\cos x\sin x+\frac{3}{4}{\cos }^{2}x+\frac{1}{4}{\sin }^{2}x+\frac{\sqrt{3}}{2}\cos x\sin x \\ =\frac{3}{2}{\cos }^{2}x+\frac{1}{2}{\sin }^{2}x \\ =\frac{3}{2}\left(1-{\sin }^{2}x\right)+\frac{1}{2}{\sin }^{2}x \\ =\frac{3}{2}-\frac{3}{2}{\sin }^{2}x+\frac{1}{2}{\sin }^{2}x \\ =\frac{3}{2}-{\sin }^{2}x \\ \mathrm{For}f\left(x\right)\mathrm{to}\mathrm{be}\mathrm{maximum},{\sin }^{2}x\text{must have minimum value, which is 0}. \\ \therefore \frac{3}{2}\mathrm{is}\mathrm{the}\mathrm{maximum}\mathrm{value}\mathrm{of}f\left(x\right). \end{gathered}$$