Step 1: Understanding the Question:
We must evaluate a trigonometric series containing the fourth powers of cosine at equally spaced angles.
Step 2: Key Formula or Approach:
We can simplify the expression using standard supplementary angle reduction formulas:
$\cos(\pi - \theta) = -\cos \theta \implies \cos^4(\pi - \theta) = \cos^4 \theta$.
And complementary angle formulas:
$\cos(\frac{\pi}{2} - \theta) = \sin \theta$.
Finally, we use the algebraic identity for powers of sine and cosine:
$\cos^4 \theta + \sin^4 \theta = 1 - 2\sin^2 \theta \cos^2 \theta = 1 - \frac{1}{2}\sin^2(2\theta)$.
Step 3: Detailed Explanation:
Let the given sum be $S$. Notice the relationships between the angles:
$$7\pi/8 = \pi - \pi/8 \implies \cos(7\pi/8) = -\cos(\pi/8) \implies \cos^4(7\pi/8) = \cos^4(\pi/8)$$
$$5\pi/8 = \pi - 3\pi/8 \implies \cos(5\pi/8) = -\cos(3\pi/8) \implies \cos^4(5\pi/8) = \cos^4(3\pi/8)$$
Substitute these back into $S$:
$$S = 2 \left[ \cos^4(\pi/8) + \cos^4(3\pi/8) \right]$$
Now, relate the remaining two angles:
$$3\pi/8 = \frac{4\pi}{8} - \frac{\pi}{8} = \frac{\pi}{2} - \frac{\pi}{8}$$
$$\cos(3\pi/8) = \cos(\frac{\pi}{2} - \frac{\pi}{8}) = \sin(\pi/8)$$
Substitute this into $S$:
$$S = 2 \left[ \cos^4(\pi/8) + \sin^4(\pi/8) \right]$$
Apply the identity $\cos^4 \theta + \sin^4 \theta = 1 - \frac{1}{2}\sin^2(2\theta)$ where $\theta = \pi/8$:
$$S = 2 \left[ 1 - \frac{1}{2}\sin^2\left(2 \times \frac{\pi}{8}\right) \right]$$
$$S = 2 \left[ 1 - \frac{1}{2}\sin^2(\pi/4) \right]$$
Since $\sin(\pi/4) = \frac{1}{\sqrt{2}}$, we have $\sin^2(\pi/4) = \frac{1}{2}$.
$$S = 2 \left[ 1 - \frac{1}{2} \left( \frac{1}{2} \right) \right] = 2 \left[ 1 - \frac{1}{4} \right] = 2 \left[ \frac{3}{4} \right] = \frac{3}{2}$$
Step 4: Final Answer:
The sum evaluates to $3/2$, which matches option (b).