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(C) Let $	heta = \frac{\pi}{5} = 36^\circ$. Then $5	heta = 180^\circ$, so $2	heta = 180^\circ - 3	heta$.Taking $\cos$ on both sides: $\cos(2	heta

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$\frac{1 + \sqrt{5}}{4}$

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(C) Let $	heta = \frac{\pi}{5} = 36^\circ$. Then $5	heta = 180^\circ$, so $2	heta = 180^\circ - 3	heta$.Taking $\cos$ on both sides: $\cos(2	heta) = \cos(180^\circ - 3	heta) = -\cos(3	heta)$.Using double and triple angle identities: $2\cos^2	heta - 1 = -(4\cos^3	heta - 3\cos	heta)$.$4\cos^3	heta + 2\cos^2	heta - 3\cos	heta - 1 = 0$.Since $\cos	heta 
eq -1$, we can divide by $(\cos	heta + 1)$ to get $4\cos^2	heta - 2\cos	heta - 1 = 0$.Using the quadratic formula for $\cos	heta$: $\cos	heta = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(4)(-1)}}{2(4)} = \frac{2 \pm \sqrt{4 + 16}}{8} = \frac{2 \pm \sqrt{20}}{8} = \frac{2 \pm 2\sqrt{5}}{8} = \frac{1 \pm \sqrt{5}}{4}$.Since $\frac{\pi}{5}$ is in the first quadrant, $\cos(\frac{\pi}{5}) > 0$, so $\cos(\frac{\pi}{5}) = \frac{1 + \sqrt{5}}{4}$.

The value of $\cos (\frac{\pi}{5})$ is ...

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