Question

The sum of the series $S = cos^4 \theta + cos^4 (\theta + \frac{2\pi}{n}) + cos^4 (\theta + \frac{4\pi}{n}) + --- upto n$ term is

• A. $\frac{n}{8}$
• B. $\frac{3n}{8}$
• C. $\frac{5n}{8}$
• D. $\frac{7n}{8}$

Answer 1

Answer: (B)

$\frac{3n}{8}$

Answer 2

We use the identity:

$$\cos^4 x = \frac{3}{8} + \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x.$$

Substitute $x = \theta + \frac{2k\pi}{n}$ for $k = 0, 1, 2, \dots, n-1$. Then,

$$S = \sum_{k=0}^{n-1} \cos^4\left(\theta + \frac{2k\pi}{n}\right) = \sum_{k=0}^{n-1} \left[\frac{3}{8} + \frac{1}{2}\cos\left(2\theta+ \frac{4k\pi}{n}\right) + \frac{1}{8}\cos\left(4\theta+ \frac{8k\pi}{n}\right)\right].$$

Splitting the sum:

$$S = n\cdot\frac{3}{8} + \frac{1}{2}\sum_{k=0}^{n-1}\cos\left(2\theta+ \frac{4k\pi}{n}\right) + \frac{1}{8}\sum_{k=0}^{n-1}\cos\left(4\theta+ \frac{8k\pi}{n}\right).$$

Since the angles in the cosine sums are equally spaced over a full period, both cosine summations vanish:

$$\sum_{k=0}^{n-1}\cos\left(2\theta+ \frac{4k\pi}{n}\right) = 0 \quad \text{and} \quad \sum_{k=0}^{n-1}\cos\left(4\theta+ \frac{8k\pi}{n}\right) = 0.$$

Thus,

$$S = \frac{3n}{8}.$$