Question

The maximum value of \[\sin^2 \left( \frac{2\pi}{3} + x \right) + \sin^2 \left( \frac{2\pi}{3} - x \right)\] is

पर्याय

• 1/2

• \[\frac{3}{2}\]

   

• 1/4

• 3/4

Answer 1

\[\frac{3}{2}\]

\[\frac{2\pi}{3} = 120^\circ\]

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