Question

If $A = \sin\frac{2\pi}{7} + \sin\frac{4\pi}{7} + \sin\frac{8\pi}{7}$ and $B = \cos\frac{2\pi}{7} + \cos\frac{4\pi}{7} + \cos\frac{8\pi}{7}$ then $\sqrt{A^2 + B^2}$ is equa

• A. 1
• B. $\sqrt{2}$
• C. 2
• D. $\sqrt{3}$

Answer 1

Answer: (B)

$\sqrt{2}$

Answer 2

To find $\sqrt{A^2 + B^2}$, we can recognize this expression as the magnitude of a complex number. Let $Z = B + iA$. Substituting the given expressions for $A$ and $B$: $Z = \left(\cos\frac{2\pi}{7} + \cos\frac{4\pi}{7} + \cos\frac{8\pi}{7}\right) + i\left(\sin\frac{2\pi}{7} + \sin\frac{4\pi}{7} + \sin\frac{8\pi}{7}\right)$

Group the terms using Euler's formula, $e^{i\theta} = \cos\theta + i\sin\theta$: $Z = \left(\cos\frac{2\pi}{7} + i\sin\frac{2\pi}{7}\right) + \left(\cos\frac{4\pi}{7} + i\sin\frac{4\pi}{7}\right) + \left(\cos\frac{8\pi}{7} + i\sin\frac{8\pi}{7}\right)$ $Z = e^{i\frac{2\pi}{7}} + e^{i\frac{4\pi}{7}} + e^{i\frac{8\pi}{7}}$

Let $\omega = e^{i\frac{2\pi}{7}}$. This is a primitive 7th root of unity. Then $Z = \omega^1 + \omega^2 + \omega^4$.

This sum is a Gaussian period for $p=7$. The quadratic residues modulo 7 are $1^2 \equiv 1$, $2^2 \equiv 4$, $3^2 \equiv 2$. So the set of quadratic residues is $R = \{1, 2, 4\}$. The sum $Z = \sum_{k \in R} \omega^k$ is denoted as $\eta_0$.

We know that for the 7th roots of unity, the sum of all roots is zero: $1 + \omega + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6 = 0$. Let $\eta_0 = \omega^1 + \omega^2 + \omega^4$ and $\eta_1 = \omega^3 + \omega^5 + \omega^6$. From the sum of all roots, we have $1 + \eta_0 + \eta_1 = 0$, so $\eta_0 + \eta_1 = -1$.

Next, consider the product $\eta_0 \eta_1$: $\eta_0 \eta_1 = (\omega^1 + \omega^2 + \omega^4)(\omega^3 + \omega^5 + \omega^6)$ $= \omega^{1+3} + \omega^{1+5} + \omega^{1+6} + \omega^{2+3} + \omega^{2+5} + \omega^{2+6} + \omega^{4+3} + \omega^{4+5} + \omega^{4+6}$ $= \omega^4 + \omega^6 + \omega^7 + \omega^5 + \omega^7 + \omega^8 + \omega^7 + \omega^9 + \omega^{10}$ Since $\omega^7 = 1$, we can simplify the powers of $\omega$: $\omega^7 = 1$, $\omega^8 = \omega^1$, $\omega^9 = \omega^2$, $\omega^{10} = \omega^3$. $\eta_0 \eta_1 = \omega^4 + \omega^6 + 1 + \omega^5 + 1 + \omega^1 + 1 + \omega^2 + \omega^3$ Rearranging the terms: $\eta_0 \eta_1 = 3 + (\omega^1 + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6)$ Since $\sum_{k=1}^{6} \omega^k = -1$: $\eta_0 \eta_1 = 3 + (-1) = 2$.

Now 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, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(2)}}{2(1)} = \frac{-1 \pm \sqrt{1 - 8}}{2} = \frac{-1 \pm \sqrt{-7}}{2} = \frac{-1 \pm i\sqrt{7}}{2}$.

So, $\{\eta_0, \eta_1\} = \left\{\frac{-1 + i\sqrt{7}}{2}, \frac{-1 - i\sqrt{7}}{2}\right\}$. To determine which value corresponds to $\eta_0 = Z = B+iA$, we need to check the sign of its imaginary part $A$. $A = \sin\frac{2\pi}{7} + \sin\frac{4\pi}{7} + \sin\frac{8\pi}{7}$. Note that $\sin\frac{8\pi}{7} = \sin(\pi + \frac{\pi}{7}) = -\sin\frac{\pi}{7}$. So, $A = \sin\frac{2\pi}{7} + \sin\frac{4\pi}{7} - \sin\frac{\pi}{7}$. Also, $\sin\frac{4\pi}{7} = \sin(\pi - \frac{4\pi}{7}) = \sin\frac{3\pi}{7}$. $A = \sin\frac{2\pi}{7} + \sin\frac{3\pi}{7} - \sin\frac{\pi}{7}$. Let $x = \frac{\pi}{7}$. Then $A = \sin(2x) + \sin(3x) - \sin(x)$. $A = 2\sin x \cos x + (3\sin x - 4\sin^3 x) - \sin x$ $A = \sin x (2\cos x + 3 - 4\sin^2 x - 1)$ $A = \sin x (2\cos x + 2 - 4(1-\cos^2 x))$ $A = \sin x (2\cos x + 2 - 4 + 4\cos^2 x)$ $A = \sin x (4\cos^2 x + 2\cos x - 2)$ $A = 2\sin x (2\cos^2 x + \cos x - 1)$ $A = 2\sin x (2\cos x - 1)(\cos x + 1)$. For $x = \frac{\pi}{7}$, $x \in (0, \frac{\pi}{2})$, so $\sin x > 0$ and $\cos x > 0$. Also, $\cos x + 1 > 0$. Since $\frac{\pi}{7} < \frac{\pi}{3}$, and $\cos x$ is a decreasing function in $(0, \pi)$, we have $\cos(\frac{\pi}{7}) > \cos(\frac{\pi}{3}) = \frac{1}{2}$. Therefore, $2\cos(\frac{\pi}{7}) > 1$, which means $2\cos(\frac{\pi}{7}) - 1 > 0$. Thus, $A > 0$.

Since $A$ is the imaginary part of $Z = \eta_0$, and $A > 0$, we must choose the root with the positive imaginary part: $Z = \eta_0 = \frac{-1 + i\sqrt{7}}{2}$. So, $B = \text{Re}(Z) = -\frac{1}{2}$ and $A = \text{Im}(Z) = \frac{\sqrt{7}}{2}$.

The problem asks for $\sqrt{A^2 + B^2}$, which is the magnitude of the complex number $Z$. $\sqrt{A^2 + B^2} = |Z| = \left|\frac{-1 + i\sqrt{7}}{2}\right|$ $|Z| = \frac{1}{2} \sqrt{(-1)^2 + (\sqrt{7})^2}$ $|Z| = \frac{1}{2} \sqrt{1 + 7}$ $|Z| = \frac{1}{2} \sqrt{8}$ $|Z| = \frac{1}{2} (2\sqrt{2})$ $|Z| = \sqrt{2}$.