Question

Find the value of $\cos\frac{2\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}$

• A. $\frac{\sqrt{13}-1}{4}$

Answer 1

Answer: (A)

$\frac{\sqrt{13}-1}{4}$

Answer 2

The problem asks for the value of the sum $\cos\frac{2\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}$.

Let $\theta = \frac{2\pi}{13}$. The sum can be written as $S = \cos\theta + \cos3\theta + \cos4\theta$.

We know that for a prime $n$, the sum of the $n$-th roots of unity is zero. $1 + e^{i\frac{2\pi}{n}} + e^{i\frac{4\pi}{n}} + \dots + e^{i\frac{2(n-1)\pi}{n}} = 0$. Taking the real part, we get: $1 + \cos\frac{2\pi}{n} + \cos\frac{4\pi}{n} + \dots + \cos\frac{2(n-1)\pi}{n} = 0$. For $n=13$: $1 + \sum_{k=1}^{12} \cos\frac{2k\pi}{13} = 0$. So, $\sum_{k=1}^{12} \cos\frac{2k\pi}{13} = -1$.

Notice the symmetry property of cosine: $\cos(2\pi - x) = \cos x$. $\cos\frac{2k\pi}{13} = \cos\left(2\pi - \frac{2k\pi}{13}\right) = \cos\left(\frac{(26-2k)\pi}{13}\right) = \cos\left(\frac{2(13-k)\pi}{13}\right)$. This means: $\cos\frac{2\pi}{13} = \cos\frac{24\pi}{13}$ $\cos\frac{4\pi}{13} = \cos\frac{22\pi}{13}$ $\cos\frac{6\pi}{13} = \cos\frac{20\pi}{13}$ $\cos\frac{8\pi}{13} = \cos\frac{18\pi}{13}$ $\cos\frac{10\pi}{13} = \cos\frac{16\pi}{13}$ $\cos\frac{12\pi}{13} = \cos\frac{14\pi}{13}$

Using this symmetry, we can rewrite the sum $\sum_{k=1}^{12} \cos\frac{2k\pi}{13}$: $2\left(\cos\frac{2\pi}{13}+\cos\frac{4\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}+\cos\frac{10\pi}{13}+\cos\frac{12\pi}{13}\right) = -1$. Let $C = \cos\frac{2\pi}{13}+\cos\frac{4\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}+\cos\frac{10\pi}{13}+\cos\frac{12\pi}{13}$. So, $C = -\frac{1}{2}$.

Let the given sum be $S_1 = \cos\frac{2\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}$. Let the remaining terms in $C$ be $S_2 = \cos\frac{4\pi}{13}+\cos\frac{10\pi}{13}+\cos\frac{12\pi}{13}$. Then $S_1 + S_2 = C = -\frac{1}{2}$.

Now, let's analyze $S_1$ and $S_2$ using the property $\cos(\pi - x) = -\cos x$: $S_1 = \cos\frac{2\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}$. Notice that $\frac{8\pi}{13} = \pi - \frac{5\pi}{13}$, so $\cos\frac{8\pi}{13} = -\cos\frac{5\pi}{13}$. $S_1 = \cos\frac{2\pi}{13}+\cos\frac{6\pi}{13}-\cos\frac{5\pi}{13}$.

For $S_2$: $\cos\frac{10\pi}{13} = \cos\left(\pi - \frac{3\pi}{13}\right) = -\cos\frac{3\pi}{13}$. $\cos\frac{12\pi}{13} = \cos\left(\pi - \frac{\pi}{13}\right) = -\cos\frac{\pi}{13}$. So, $S_2 = \cos\frac{4\pi}{13} - \cos\frac{3\pi}{13} - \cos\frac{\pi}{13}$.

This looks like a problem related to Gaussian periods. For a prime $p$, the sum of cosines of angles $2k\pi/p$ where $k$ are quadratic residues or non-residues modulo $p$ can be evaluated. For $p=13$: Quadratic residues modulo 13 are $Q = \{1^2, 2^2, 3^2, 4^2, 5^2, 6^2\} \pmod{13} = \{1, 4, 9, 3, 12, 10\}$. In increasing order: $Q = \{1, 3, 4, 9, 10, 12\}$. Quadratic non-residues modulo 13 are $N = \{2, 5, 6, 7, 8, 11\}$.

Consider the sum $\eta_0 = \sum_{k \in Q} e^{i2k\pi/13}$. Its real part is $\text{Re}(\eta_0) = \sum_{k \in Q} \cos(2k\pi/13)$. $\text{Re}(\eta_0) = \cos\frac{2\pi}{13} + \cos\frac{6\pi}{13} + \cos\frac{8\pi}{13} + \cos\frac{18\pi}{13} + \cos\frac{20\pi}{13} + \cos\frac{24\pi}{13}$. Using symmetry $\cos\frac{2k\pi}{13} = \cos\frac{2(13-k)\pi}{13}$: $\cos\frac{18\pi}{13} = \cos\frac{2(13-9)\pi}{13} = \cos\frac{8\pi}{13}$. $\cos\frac{20\pi}{13} = \cos\frac{2(13-10)\pi}{13} = \cos\frac{6\pi}{13}$. $\cos\frac{24\pi}{13} = \cos\frac{2(13-12)\pi}{13} = \cos\frac{2\pi}{13}$. So, $\text{Re}(\eta_0) = 2\left(\cos\frac{2\pi}{13} + \cos\frac{6\pi}{13} + \cos\frac{8\pi}{13}\right) = 2S_1$.

Consider the sum $\eta_1 = \sum_{k \in N} e^{i2k\pi/13}$. Its real part is $\text{Re}(\eta_1) = \sum_{k \in N} \cos(2k\pi/13)$. $\text{Re}(\eta_1) = \cos\frac{4\pi}{13} + \cos\frac{10\pi}{13} + \cos\frac{12\pi}{13} + \cos\frac{14\pi}{13} + \cos\frac{16\pi}{13} + \cos\frac{22\pi}{13}$. Using symmetry: $\cos\frac{14\pi}{13} = \cos\frac{12\pi}{13}$. $\cos\frac{16\pi}{13} = \cos\frac{10\pi}{13}$. $\cos\frac{22\pi}{13} = \cos\frac{4\pi}{13}$. So, $\text{Re}(\eta_1) = 2\left(\cos\frac{4\pi}{13} + \cos\frac{10\pi}{13} + \cos\frac{12\pi}{13}\right) = 2S_2$.

We know that $1 + \sum_{k=1}^{12} \cos\frac{2k\pi}{13} = 0$, which implies $1 + \text{Re}(\eta_0) + \text{Re}(\eta_1) = 0$. So, $1 + 2S_1 + 2S_2 = 0$, which means $2(S_1+S_2) = -1$, or $S_1+S_2 = -1/2$. This matches our earlier result.

For a prime $p \equiv 1 \pmod 4$, the Gaussian periods are given by: $2\text{Re}(\eta_0) = -1 + \sqrt{p}$ $2\text{Re}(\eta_1) = -1 - \sqrt{p}$ Since $13 \equiv 1 \pmod 4$, we can use these formulas. We are looking for $S_1$. We found that $2S_1 = \text{Re}(\eta_0)$. So, $2S_1 = \frac{-1 + \sqrt{13}}{2}$. Therefore, $S_1 = \frac{-1 + \sqrt{13}}{4}$.

The final answer is $\boxed{\frac{\sqrt{13}-1}{4}}$.

Explanation of the solution:

1. Utilize Sum of Roots of Unity: The sum of the $n$-th roots of unity is zero. For $z^{13}-1=0$, the roots are $e^{i\frac{2k\pi}{13}}$ for $k=0, 1, \dots, 12$. The sum of these roots is zero. Taking the real part, we get $1 + \sum_{k=1}^{12} \cos\frac{2k\pi}{13} = 0$, which implies $\sum_{k=1}^{12} \cos\frac{2k\pi}{13} = -1$.
2. Exploit Cosine Symmetry: Use the property $\cos(2\pi - x) = \cos x$ to show that $\cos\frac{2k\pi}{13} = \cos\frac{2(13-k)\pi}{13}$. This means the terms in the sum repeat, allowing us to write $2\left(\cos\frac{2\pi}{13}+\cos\frac{4\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}+\cos\frac{10\pi}{13}+\cos\frac{12\pi}{13}\right) = -1$. Let $C_P = \cos\frac{2\pi}{13}+\cos\frac{4\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}+\cos\frac{10\pi}{13}+\cos\frac{12\pi}{13}$. So $C_P = -1/2$.
3. Connect to Gaussian Periods: The given sum $\cos\frac{2\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}$ corresponds to the terms whose indices $k$ are quadratic residues modulo 13 (i.e., $1, 3, 4$). The quadratic residues modulo 13 are $Q = \{1, 3, 4, 9, 10, 12\}$. The sum of cosines for these indices is $\sum_{k \in Q} \cos\frac{2k\pi}{13} = \cos\frac{2\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}+\cos\frac{18\pi}{13}+\cos\frac{20\pi}{13}+\cos\frac{24\pi}{13}$. Using symmetry again, this sum is $2(\cos\frac{2\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13})$, which is $2S_1$. This sum is the real part of the Gaussian period $\eta_0 = \sum_{k \in Q} e^{i2k\pi/13}$.
4. Apply Formula for Gaussian Periods: For a prime $p \equiv 1 \pmod 4$, the real part of the Gaussian period $\eta_0$ is given by $\text{Re}(\eta_0) = \frac{-1 + \sqrt{p}}{2}$. Since $13 \equiv 1 \pmod 4$, we have $\text{Re}(\eta_0) = \frac{-1 + \sqrt{13}}{2}$. As $2S_1 = \text{Re}(\eta_0)$, we have $2S_1 = \frac{-1 + \sqrt{13}}{2}$. Therefore, $S_1 = \frac{-1 + \sqrt{13}}{4}$.

Answer:

The value of $\cos\frac{2\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}$ is $\frac{\sqrt{13}-1}{4}$.