cos ^3 theta + cos ^3(120^{circ} + theta) + c | vedclass

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(D) We use the identity $\cos 3A = 4 \cos^3 A - 3 \cos A$,which implies $\cos^3 A = \frac{1}{4} (\cos 3A + 3 \cos A)$.Applying this to each term:$\cos

Vedclass · Accepted Answer

$\frac{3}{4} \cos 3 	heta$

Vedclass · Answer

(D) We use the identity $\cos 3A = 4 \cos^3 A - 3 \cos A$,which implies $\cos^3 A = \frac{1}{4} (\cos 3A + 3 \cos A)$.Applying this to each term:$\cos^3 	heta = \frac{1}{4} (\cos 3	heta + 3 \cos 	heta)$$\cos^3(120^{\circ} + 	heta) = \frac{1}{4} (\cos(360^{\circ} + 3	heta) + 3 \cos(120^{\circ} + 	heta)) = \frac{1}{4} (\cos 3	heta + 3 \cos(120^{\circ} + 	heta))$$\cos^3(	heta - 120^{\circ}) = \frac{1}{4} (\cos(3	heta - 360^{\circ}) + 3 \cos(	heta - 120^{\circ})) = \frac{1}{4} (\cos 3	heta + 3 \cos(	heta - 120^{\circ}))$Summing these up:Sum $= \frac{1}{4} [3 \cos 3	heta + 3 (\cos 	heta + \cos(120^{\circ} + 	heta) + \cos(	heta - 120^{\circ}))]$Using the sum-to-product formula $\cos(A+B) + \cos(A-B) = 2 \cos A \cos B$:$\cos(120^{\circ} + 	heta) + \cos(	heta - 120^{\circ}) = 2 \cos 	heta \cos 120^{\circ} = 2 \cos 	heta (-1/2) = -\cos 	heta$Substituting this back:Sum $= \frac{1}{4} [3 \cos 3	heta + 3 (\cos 	heta - \cos 	heta)] = \frac{3}{4} \cos 3	heta$.

$\cos ^3 \theta + \cos ^3(120^{\circ} + \theta) + \cos ^3(\theta - 120^{\circ}) = $

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