The period of sin(π3−x)sin(π3+x) is

sin(π3−x)sin(π3+x)=2sin(π3−x).sin(π3+x)2 =cos(π3−x−π3−x)−cos(π3−x+π3+x)2 =cos(−2x)−cos(π2+30°)2 =cos(−2x)−sin30°2 ∴ Period of cos(−2x)=2π|−2|=π

sin(π3−x)sin(π3+x)=2sin(π3−x).sin(π3+x)2 =cos(π3−x−π3−x)−cos(π3−x+π3+x)2 =cos(−2x)−cos(π2+30°)2 =cos(−2x)−sin30°2 ∴ Period of cos(−2x)=2π|−2|=π

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The period of $\sin (\frac{π}{3} - x) \sin (\frac{π}{3} + x)$ is

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\frac{π}{2}

π

\frac{3 π}{2}

2 π

answer is B.

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\sin (\frac{π}{3} - x) \sin (\frac{π}{3} + x) = \frac{2 \sin (\frac{π}{3} - x) . \sin (\frac{π}{3} + x)}{2}

$= \frac{\cos (\frac{π}{3} - x - \frac{π}{3} - x) - \cos (\frac{π}{3} - x + \frac{π}{3} + x)}{2}$

$= \frac{\cos (- 2 x) - \cos (\frac{π}{2} + 30 °)}{2}$

$= \frac{\cos (- 2 x) - \sin 30 °}{2}$

$∴$ Period of $\cos (- 2 x) = \frac{2 π}{| - 2 |} = π$

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