cos 12^{circ} cdot cos 24^{circ} cdot cos 36^ | vedclass

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Question

(C) We use the identity $\cos 	heta \cos(60^{\circ}-	heta) \cos(60^{\circ}+	heta) = \frac{1}{4} \cos 3	heta$.The given expression is $P = (\cos 12

Vedclass · Accepted Answer

$\frac{1}{64}$

Vedclass · Answer

(C) We use the identity $\cos 	heta \cos(60^{\circ}-	heta) \cos(60^{\circ}+	heta) = \frac{1}{4} \cos 3	heta$.The given expression is $P = (\cos 12^{\circ} \cos 48^{\circ} \cos 84^{\circ}) \cdot (\cos 24^{\circ} \cos 72^{\circ} \cos 36^{\circ})$.For the first part: $\cos 12^{\circ} \cos(60^{\circ}-12^{\circ}) \cos(60^{\circ}+12^{\circ}) = \frac{1}{4} \cos(3 	imes 12^{\circ}) = \frac{1}{4} \cos 36^{\circ}$.For the second part: $\cos 24^{\circ} \cos(60^{\circ}-24^{\circ}) \cos(60^{\circ}+24^{\circ}) = \frac{1}{4} \cos(3 	imes 24^{\circ}) = \frac{1}{4} \cos 72^{\circ}$.Thus,$P = (\frac{1}{4} \cos 36^{\circ}) \cdot (\frac{1}{4} \cos 72^{\circ}) = \frac{1}{16} \cos 36^{\circ} \cos 72^{\circ}$.Using $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$ and $\cos 72^{\circ} = \frac{\sqrt{5}-1}{4}$,we get:$P = \frac{1}{16} \left( \frac{\sqrt{5}+1}{4} ight) \left( \frac{\sqrt{5}-1}{4} ight) = \frac{1}{16} \left( \frac{5-1}{16} ight) = \frac{1}{16} \cdot \frac{4}{16} = \frac{1}{64}$.

$\cos 12^{\circ} \cdot \cos 24^{\circ} \cdot \cos 36^{\circ} \cdot \cos 48^{\circ} \cdot \cos 72^{\circ} \cdot \cos 84^{\circ} = $

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