(4 cos^2 9^{circ} - 3)(4 cos^2 27^{circ} - 3) | vedclass

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Question

(A) We use the identity $\cos 3	heta = 4 \cos^3 	heta - 3 \cos 	heta = \cos 	heta (4 \cos^2 	heta - 3)$.Thus,$4 \cos^2 	heta - 3 = \frac{\cos 3\

Vedclass · Accepted Answer

$	an 9^{\circ}$

Vedclass · Answer

(A) We use the identity $\cos 3	heta = 4 \cos^3 	heta - 3 \cos 	heta = \cos 	heta (4 \cos^2 	heta - 3)$.Thus,$4 \cos^2 	heta - 3 = \frac{\cos 3	heta}{\cos 	heta}$.Substituting this into the expression:$(4 \cos^2 9^{\circ} - 3)(4 \cos^2 27^{\circ} - 3) = \left( \frac{\cos 27^{\circ}}{\cos 9^{\circ}} ight) \left( \frac{\cos 81^{\circ}}{\cos 27^{\circ}} ight)$$= \frac{\cos 81^{\circ}}{\cos 9^{\circ}}$$= \frac{\cos(90^{\circ} - 9^{\circ})}{\cos 9^{\circ}}$$= \frac{\sin 9^{\circ}}{\cos 9^{\circ}} = 	an 9^{\circ}$.

$(4 \cos^2 9^{\circ} - 3)(4 \cos^2 27^{\circ} - 3) = $

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