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प्रश्न
Prove that: \[\cos 6° \cos 42° \cos 66° \cos 78° = \frac{1}{16}\]
उत्तर
\[LHS = \cos6° \cos42° \cos66° \cos78° \]
\[ = \frac{1}{4}\left( 2\cos6° \cos66° \right)\left( 2\cos42° \cos78° \right) \]
\[ = \frac{1}{4}\left( \cos72° + \cos60° \right)\left( \cos120° + \cos36° \right) \left[ \because 2\text{ cos }A\text{ cos } B = \cos\left( A + B \right) + \cos\left( A - B \right) \right] \]
\[ = \frac{1}{4}\left\{ \cos\left( 90° - 72° \right) + \frac{1}{2} \right\}\left\{ - \frac{1}{2} + \frac{\sqrt{5} + 1}{4} \right\}\]
\[= \frac{1}{4}\left( \sin18° + \frac{1}{2} \right)\left( - \frac{1}{2} + \frac{\sqrt{5} + 1}{4} \right)\]
\[ = \frac{1}{4}\left( \frac{\sqrt{5} - 1}{4} + \frac{1}{2} \right)\left( \frac{\sqrt{5} + 1}{4} - \frac{1}{2} \right)\]
\[ = \frac{1}{4}\left( \frac{\sqrt{5} - 1 + 2}{4} \right)\left( \frac{\sqrt{5} + 1 - 2}{4} \right)\]
\[ = \frac{1}{64}\left( \sqrt{5} + 1 \right)\left( \sqrt{5} - 1 \right)\]
\[ = \frac{1}{64}\left( 5 - 1 \right)\]
\[ = \frac{1}{16} = RHS\]
\[\text{ Hence proved } .\]
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