If f(theta) = cos^3 theta + cos^3 left(frac{2 | vedclass

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(A) We use the identity $\cos^3 \alpha + \cos^3 \left(\alpha + \frac{2\pi}{3}\right) + \cos^3 \left(\alpha + \frac{4\pi}{3}\right) = \frac{3}{4} \cos(

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$\frac{3(\sqrt{5}-1)}{16}$

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(A) We use the identity $\cos^3 \alpha + \cos^3 \left(\alpha + \frac{2\pi}{3}ight) + \cos^3 \left(\alpha + \frac{4\pi}{3}ight) = \frac{3}{4} \cos(3\alpha)$.Given $f(	heta) = \cos^3 	heta + \cos^3 \left(	heta + \frac{2\pi}{3}ight) + \cos^3 \left(	heta - \frac{2\pi}{3}ight)$.Since $\cos \left(	heta - \frac{2\pi}{3}ight) = \cos \left(	heta - \frac{2\pi}{3} + 2\piight) = \cos \left(	heta + \frac{4\pi}{3}ight)$,the expression simplifies to $f(	heta) = \frac{3}{4} \cos(3	heta)$.Substituting $	heta = \frac{\pi}{5}$,we get $f\left(\frac{\pi}{5}ight) = \frac{3}{4} \cos \left(\frac{3\pi}{5}ight)$.We know $\cos \left(\frac{3\pi}{5}ight) = -\sin \left(\frac{\pi}{10}ight) = -\left(\frac{\sqrt{5}-1}{4}ight) = \frac{1-\sqrt{5}}{4}$.Wait,$\cos \left(\frac{3\pi}{5}ight) = \cos(108^\circ) = \frac{1-\sqrt{5}}{4}$.Thus,$f\left(\frac{\pi}{5}ight) = \frac{3}{4} \left(\frac{1-\sqrt{5}}{4}ight) = \frac{3(1-\sqrt{5})}{16}$.

If $f(\theta) = \cos^3 \theta + \cos^3 \left(\frac{2\pi}{3} + \theta\right) + \cos^3 \left(\theta - \frac{2\pi}{3}\right)$,then $f\left(\frac{\pi}{5}\right) = $

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