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# The Maximum Value of Sin 2 ( 2 π 3 + X ) + Sin 2 ( 2 π 3 − X ) is - CBSE (Science) Class 11 - Mathematics

ConceptTrigonometric Functions of Sum and Difference of Two Angles

#### Question

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

• 1/2

• $\frac{3}{2}$

• 1/4

• 3/4

#### Solution

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

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#### APPEARS IN

RD Sharma Solution for Mathematics Class 11 (2019 to Current)
Chapter 7: Values of Trigonometric function at sum or difference of angles
Q: 20 | Page no. 28

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Solution The Maximum Value of Sin 2 ( 2 π 3 + X ) + Sin 2 ( 2 π 3 − X ) is Concept: Trigonometric Functions of Sum and Difference of Two Angles.
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