Question

$\sin(\frac{2\pi}{7})+\sin(\frac{4\pi}{7})+\sin(\frac{8\pi}{7})$

Answer 1

$\frac{\sqrt{7}}{2}$

Answer 2

The problem asks to evaluate the sum $S = \sin(\frac{2\pi}{7})+\sin(\frac{4\pi}{7})+\sin(\frac{8\pi}{7})$.

First, let's simplify the term $\sin(\frac{8\pi}{7})$: $\sin(\frac{8\pi}{7}) = \sin(\pi + \frac{\pi}{7}) = -\sin(\frac{\pi}{7})$.

So the sum becomes: $S = \sin(\frac{2\pi}{7})+\sin(\frac{4\pi}{7})-\sin(\frac{\pi}{7})$.

This is a common problem that can be solved using properties of complex numbers and roots of unity, specifically Gaussian periods. Let $\omega = e^{i\frac{2\pi}{7}}$. The roots of $z^7 - 1 = 0$ are $\omega^k$ for $k=0, 1, \dots, 6$. The sum of all roots is zero: $\sum_{k=0}^{6} \omega^k = 1 + \omega + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6 = 0$.

Let's consider the quadratic residues modulo 7. The non-zero residues are $\{1, 2, 3, 4, 5, 6\}$. The quadratic residues are $1^2 \equiv 1$, $2^2 \equiv 4$, $3^2 \equiv 9 \equiv 2$. So, the set of quadratic residues is $R = \{1, 2, 4\}$. The set of quadratic non-residues is $N = \{3, 5, 6\}$.

Define the Gaussian periods: $\eta_0 = \sum_{k \in R} \omega^k = \omega^1 + \omega^2 + \omega^4 = e^{i\frac{2\pi}{7}} + e^{i\frac{4\pi}{7}} + e^{i\frac{8\pi}{7}}$. $\eta_1 = \sum_{k \in N} \omega^k = \omega^3 + \omega^5 + \omega^6 = e^{i\frac{6\pi}{7}} + e^{i\frac{10\pi}{7}} + e^{i\frac{12\pi}{7}}$.

From the sum of all roots: $1 + \eta_0 + \eta_1 = 0 \implies \eta_0 + \eta_1 = -1$.

Now, let's consider the product $\eta_0 \eta_1$: $\eta_0 \eta_1 = (\omega^1 + \omega^2 + \omega^4)(\omega^3 + \omega^5 + \omega^6)$ $= \omega^4 + \omega^6 + \omega^7 + \omega^5 + \omega^7 + \omega^8 + \omega^7 + \omega^9 + \omega^{10}$. Since $\omega^7 = 1$, we have: $\eta_0 \eta_1 = \omega^4 + \omega^6 + 1 + \omega^5 + 1 + \omega^1 + 1 + \omega^2 + \omega^3$ (using $\omega^8=\omega$, $\omega^9=\omega^2$, $\omega^{10}=\omega^3$) Rearranging the terms: $\eta_0 \eta_1 = 3 + (\omega^1 + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6)$. Since $\sum_{k=1}^{6} \omega^k = -1$: $\eta_0 \eta_1 = 3 + (-1) = 2$.

So we have a system of equations for $\eta_0$ and $\eta_1$:

1. $\eta_0 + \eta_1 = -1$
2. $\eta_0 \eta_1 = 2$

$\eta_0$ and $\eta_1$ are the roots of the quadratic equation $x^2 - (\eta_0 + \eta_1)x + \eta_0 \eta_1 = 0$: $x^2 - (-1)x + 2 = 0 \implies x^2 + x + 2 = 0$. Using the quadratic formula, $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(2)}}{2(1)} = \frac{-1 \pm \sqrt{1 - 8}}{2} = \frac{-1 \pm \sqrt{-7}}{2} = \frac{-1 \pm i\sqrt{7}}{2}$.

So, $\{\eta_0, \eta_1\} = \{\frac{-1 + i\sqrt{7}}{2}, \frac{-1 - i\sqrt{7}}{2}\}$. We need to determine which one is $\eta_0$. $\eta_0 = \cos(\frac{2\pi}{7}) + \cos(\frac{4\pi}{7}) + \cos(\frac{8\pi}{7}) + i(\sin(\frac{2\pi}{7}) + \sin(\frac{4\pi}{7}) + \sin(\frac{8\pi}{7}))$. The sum we want to evaluate is the imaginary part of $\eta_0$.

Let's analyze the real part of $\eta_0$: $\text{Re}(\eta_0) = \cos(\frac{2\pi}{7}) + \cos(\frac{4\pi}{7}) + \cos(\frac{8\pi}{7})$. We know $\cos(\frac{8\pi}{7}) = \cos(2\pi - \frac{6\pi}{7}) = \cos(\frac{6\pi}{7})$. So, $\text{Re}(\eta_0) = \cos(\frac{2\pi}{7}) + \cos(\frac{4\pi}{7}) + \cos(\frac{6\pi}{7})$. We know that $\cos(\frac{2\pi}{7}) > 0$, $\cos(\frac{4\pi}{7}) < 0$, $\cos(\frac{6\pi}{7}) < 0$. However, the sum $\cos(\frac{2\pi}{7}) + \cos(\frac{4\pi}{7}) + \cos(\frac{6\pi}{7})$ is negative. Consider the full sum of cosines: $\sum_{k=1}^{6} \cos(\frac{2k\pi}{7}) = \cos(\frac{2\pi}{7}) + \cos(\frac{4\pi}{7}) + \cos(\frac{6\pi}{7}) + \cos(\frac{8\pi}{7}) + \cos(\frac{10\pi}{7}) + \cos(\frac{12\pi}{7}) = -1$. Using $\cos(\frac{2k\pi}{7}) = \cos(\frac{2(7-k)\pi}{7})$, we have: $\cos(\frac{8\pi}{7}) = \cos(\frac{6\pi}{7})$ $\cos(\frac{10\pi}{7}) = \cos(\frac{4\pi}{7})$ $\cos(\frac{12\pi}{7}) = \cos(\frac{2\pi}{7})$ So, $2(\cos(\frac{2\pi}{7}) + \cos(\frac{4\pi}{7}) + \cos(\frac{6\pi}{7})) = -1$. Therefore, $\cos(\frac{2\pi}{7}) + \cos(\frac{4\pi}{7}) + \cos(\frac{6\pi}{7}) = -\frac{1}{2}$. This means $\text{Re}(\eta_0) = -\frac{1}{2}$.

Comparing this with the possible values for $x$: $\frac{-1 + i\sqrt{7}}{2}$ has real part $-\frac{1}{2}$. $\frac{-1 - i\sqrt{7}}{2}$ has real part $-\frac{1}{2}$. Both have the same real part.

Now we need to determine the sign of the imaginary part. $S = \sin(\frac{2\pi}{7})+\sin(\frac{4\pi}{7})-\sin(\frac{\pi}{7})$. We know that for $x \in (0, \pi)$, $\sin(x) > 0$. $\frac{2\pi}{7} \approx 0.897$ radians, which is in the first quadrant. $\sin(\frac{2\pi}{7}) > 0$. $\frac{4\pi}{7} \approx 1.795$ radians, which is in the second quadrant. $\sin(\frac{4\pi}{7}) > 0$. $\frac{\pi}{7} \approx 0.448$ radians, which is in the first quadrant. $\sin(\frac{\pi}{7}) > 0$.

We need to compare the magnitudes. Consider the function $f(x) = \sin(x)$. It is concave down on $(0, \pi)$. The angles are $\frac{\pi}{7}, \frac{2\pi}{7}, \frac{4\pi}{7}$. We know $\sin(\frac{4\pi}{7}) = \sin(\frac{3\pi}{7})$. So $S = \sin(\frac{2\pi}{7}) + \sin(\frac{3\pi}{7}) - \sin(\frac{\pi}{7})$. Let $x = \frac{\pi}{7}$. Then $S = \sin(2x) + \sin(3x) - \sin(x)$. $S = 2\sin(x)\cos(x) + (3\sin(x) - 4\sin^3(x)) - \sin(x)$. $S = \sin(x)(2\cos(x) + 3 - 4\sin^2(x) - 1)$. $S = \sin(x)(2\cos(x) + 2 - 4(1-\cos^2(x)))$. $S = \sin(x)(2\cos(x) + 2 - 4 + 4\cos^2(x))$. $S = \sin(x)(4\cos^2(x) + 2\cos(x) - 2)$. $S = 2\sin(x)(2\cos^2(x) + \cos(x) - 1)$. $S = 2\sin(x)(2\cos(x)-1)(\cos(x)+1)$. Since $x = \frac{\pi}{7}$, which is in the first quadrant, $\sin(x) > 0$ and $\cos(x) > 0$. Also, $\cos(x)+1 > 0$. We need to check the sign of $(2\cos(x)-1)$. For $x = \frac{\pi}{7}$, $x < \frac{\pi}{3}$. Since $\cos(x)$ is a decreasing function in $(0, \pi)$, $\cos(\frac{\pi}{7}) > \cos(\frac{\pi}{3}) = \frac{1}{2}$. So $2\cos(\frac{\pi}{7}) > 1$, which means $2\cos(\frac{\pi}{7})-1 > 0$. Therefore, $S > 0$.

Since $\text{Re}(\eta_0) = -\frac{1}{2}$ and $\text{Im}(\eta_0) > 0$, we must have: $\eta_0 = \frac{-1 + i\sqrt{7}}{2}$.

The value we are looking for is the imaginary part of $\eta_0$. $\text{Im}(\eta_0) = \frac{\sqrt{7}}{2}$.