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प्रश्न
Prove that \[\left| \cos x \cos \left( \frac{\pi}{3} - x \right) \cos \left( \frac{\pi}{3} + x \right) \right| \leq \frac{1}{4}\] for all values of x
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उत्तर
\[\frac{\pi}{3} = 60°\]
\[\text{ We have } , \]
\[\left| \text{ cos } x \cos\left( 60° - x \right) \cos\left( 60° + x \right) \right|\]
\[ = \left| \text{ cos } x\left( \cos^2 60° - \sin^2 x \right) \right| \left[ \cos^2 A - \sin^2 B = \cos\left( A - B \right) \cos\left( A + B \right) \right]\]
\[ = \left| \text{ cos } x\left( \frac{1}{4} - \sin^2 x \right) \right| \]
\[ = \left| \text{ cos } x\frac{1}{4}\left( 1 - 4 \sin^2 x \right) \right|\]
\[ = \left| \frac{1}{4}\text{ cos } x\left\{ 1 - 4\left( 1 - \cos^2 x \right) \right\} \right|\]
\[ = \left| \frac{1}{4}\text{ cos } x\left\{ - 3 + 4 \cos^2 x \right\} \right|\]
\[= \left| \frac{1}{4}\left( 4 \cos^3 x - 3\text{ cos } x \right) \right|\]
\[ = \left| \frac{1}{4}\cos3x \right| \left[ \because \text{ cos } 3x = 4 \cos^3 x - 3\text{ cos } x \right] \]
\[ \leq \frac{1}{4} \left( \because \left| \text{ cos } x \right| \leq 1 \text{ for all } x \right)\]
\[ \therefore \left| \text{ cos } x \cos\left( 60°- x \right) \cos\left( 60° + x \right) \right| \leq \frac{1}{4}\]
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