Maximum and minimum values of sin2⁡120∘+θ+sin2⁡120∘−θ are

sin2⁡120∘+θ+sin2⁡120∘−θ=1−cos2⁡120∘+θ+sin2⁡120∘−θ=1−cos2⁡120∘+θ−sin2⁡120∘−θ=1−cos⁡240∘cos⁡2θ=1+12cos⁡2θ∴Maximum value =1+12(1)=32, Minimum value =1+12(−1)=12.

Maximum and minimum values of sin2⁡120∘+θ+sin2⁡120∘−θ are

sin2⁡120∘+θ+sin2⁡120∘−θ=1−cos2⁡120∘+θ+sin2⁡120∘−θ=1−cos2⁡120∘+θ−sin2⁡120∘−θ=1−cos⁡240∘cos⁡2θ=1+12cos⁡2θ∴Maximum value =1+12(1)=32, Minimum value =1+12(−1)=12.

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Maximum and minimum values of $\sin^{2} (120^{\circ} + θ) + \sin^{2} (120^{\circ} - θ) are$

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max = 3/2, min = 1/2

max = 1/2, min = 0

max = 3/2, min = 0

max = 3/2, min = 1/3

answer is A.

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Detailed Solution

$\begin{array}{l} \sin^{2} (120^{\circ} + θ) + \sin^{2} (120^{\circ} - θ) = 1 - \cos^{2} (120^{\circ} + θ) + \sin^{2} (120^{\circ} - θ) \\ = 1 - [\cos^{2} (120^{\circ} + θ) - \sin^{2} (120^{\circ} - θ)] = 1 - \cos 240^{\circ} \cos 2 θ = 1 + \frac{1}{2} \cos 2 θ \end{array}$
$∴$ Maximum value $= 1 + \frac{1}{2} (1) = \frac{3}{2}$ , Minimum value $= 1 + \frac{1}{2} (- 1) = \frac{1}{2}$ .