cos⁡6∘sin⁡24∘cos⁡72∘=

=122sin⁡24∘cos⁡6∘cos⁡72∘=12sin⁡30∘+sin⁡18∘cos⁡72∘=1212+5−145−14=18.

cos⁡6∘sin⁡24∘cos⁡72∘=

=122sin⁡24∘cos⁡6∘cos⁡72∘=12sin⁡30∘+sin⁡18∘cos⁡72∘=1212+5−145−14=18.

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$\cos 6^{\circ} \sin 24^{\circ} \cos 72^{\circ} =$

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1/4

answer is C.

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Detailed Solution

$\begin{array}{l} = \frac{1}{2} (2 \sin 24^{\circ} \cos 6^{\circ}) \cos 72^{\circ} = \frac{1}{2} (\sin 30^{\circ} + \sin 18^{\circ}) \cos 72^{\circ} \\ = \frac{1}{2} [\frac{1}{2} + \frac{\sqrt{5} - 1}{4}] (\frac{\sqrt{5} - 1}{4}) = \frac{1}{8} . \end{array}$