The maximum value of sin

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

Solution:

(b) 3/2

$\frac{2 \pi}{3}=120^{\circ}$

Let $f(x)=\sin ^{2}(90+30+x)+\sin ^{2}(90+30-x)$

$=[\cos (30+x)]^{2}+[\cos (30-x)]^{2} \quad[$ Using $\sin (90+A)=\cos A]$

$=\left[\frac{\sqrt{3}}{2} \cos x-\frac{1}{2} \sin x\right]^{2}+\left[\frac{\sqrt{3}}{2} \cos x+\frac{1}{2} \sin x\right]^{2}$

$=\frac{3}{4} \cos ^{2} x+\frac{1}{4} \sin ^{2} x-\frac{\sqrt{3}}{2} \cos x \sin x+\frac{3}{4} \cos ^{2} x+\frac{1}{4} \sin ^{2} x+\frac{\sqrt{3}}{2} \cos x \sin x$

$=\frac{3}{2} \cos ^{2} x+\frac{1}{2} \sin ^{2} x$

$=\frac{3}{2}\left(1-\sin ^{2} x\right)+\frac{1}{2} \sin ^{2} x$

$=\frac{3}{2}-\frac{3}{2} \sin ^{2} x+\frac{1}{2} \sin ^{2} x$

$=\frac{3}{2}-\sin ^{2} x$

For $f(x)$ to be maximum, $\sin ^{2} x$ must have minimum value, which is 0 .

$\therefore \frac{3}{2}$ is the maximum value of $f(x)$.