Find the Maximum Value of sin²(2π/3 + x) + sin²(2π/3 − x)

Solution

Use the identity:

\[ \sin^2\theta=\frac{1-\cos2\theta}{2} \]

Therefore,

\[ \sin^2\left(\frac{2\pi}{3}+x\right) = \frac{ 1-\cos\left(\frac{4\pi}{3}+2x\right) }{2} \]

and

\[ \sin^2\left(\frac{2\pi}{3}-x\right) = \frac{ 1-\cos\left(\frac{4\pi}{3}-2x\right) }{2} \]

Adding,

\[ = \frac{ 2-\left[ \cos\left(\frac{4\pi}{3}+2x\right) + \cos\left(\frac{4\pi}{3}-2x\right) \right] }{2} \]

Using the identity:

\[ \cos C+\cos D = 2\cos\frac{C+D}{2}\cos\frac{C-D}{2} \]

we get

\[ = \frac{ 2-2\cos\frac{8\pi/3}{2}\cos\frac{4x}{2} }{2} \]

\[ = \frac{ 2-2\cos\frac{4\pi}{3}\cos2x }{2} \]

Since

\[ \cos\frac{4\pi}{3}=-\frac{1}{2} \]

Therefore,

\[ = \frac{ 2+\cos2x }{2} \]

\[ = 1+\frac{1}{2}\cos2x \]

Now,

\[ -1\leq\cos2x\leq1 \]

Hence the maximum value occurs when

\[ \cos2x=1 \]

Therefore,

\[ 1+\frac{1}{2}(1) = \frac{3}{2} \]

Final Answer

\[ \boxed{ \max\left[ \sin^2\left(\frac{2\pi}{3}+x\right) + \sin^2\left(\frac{2\pi}{3}-x\right) \right] = \frac{3}{2} } \]

Correct Option: (b)