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प्रश्न
Prove that \[\left| \sin x \sin \left( \frac{\pi}{3} - x \right) \sin \left( \frac{\pi}{3} + x \right) \right| \leq \frac{1}{4}\] for all values of x
उत्तर
\[\frac{\pi}{3} = 60°\]
\[\text{ We have } , \]
\[ \left| \text{ sin }x \sin\left( 60 - x \right) \sin\left( 60 + x \right) \right|\]
\[ = \left| \text{ sin } x\left( \sin^2 60 - \sin^2 x \right) \right|\]
\[ \left[ \because \sin\left( A + B \right) \sin\left( A - B \right) = \sin^2 A - \sin^2 B \right] \]
\[ = \left| \text{ sin } x\left( \frac{3}{4} - \sin^2 x \right) \right| \]
\[ = \left| \frac{1}{4}\text{ sin } x\left( 3 - 4 \sin^2 x \right) \right|\]
\[ = \left| \frac{1}{4}3\text{ sin } x - 4 \sin^3 x \right| \]
\[ = \frac{1}{4}\left| \sin3x \right| \left( \because 3\text{ sin } x - 4 \sin^3 x = \sin3x \right)\]
\[ \leq \frac{1}{4} \left( \because \left| \text{ sin } x \right| \leq 1 for all x \right)\]
\[ \therefore \left| \text{ sin } x \sin\left( 60 - x \right) \sin\left( 60 + x \right) \right| \leq \frac{1}{4}\]
\[\text{ Hence proved } .\]
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