Question

The value of $\bigg|4\bigg(2cos^3\frac{\pi}{7}-cos^2\frac{\pi}{7}-cos\frac{\pi}{7}\bigg)\bigg|$ is

Answer 1

1

Answer 2

Let $x = \frac{\pi}{7}$. The expression inside the absolute value is $E = 4(2\cos^3 x - \cos^2 x - \cos x)$.

We use the triple angle formula for cosine: $\cos(3x) = 4\cos^3 x - 3\cos x$. From this, $4\cos^3 x = \cos(3x) + 3\cos x$. Dividing by 2, $2\cos^3 x = \frac{1}{2}(\cos(3x) + 3\cos x)$.

Substitute this into the expression for $E$: $E = 4\left(\frac{1}{2}(\cos(3x) + 3\cos x) - \cos^2 x - \cos x\right)$ $E = 2(\cos(3x) + 3\cos x) - 4\cos^2 x - 4\cos x$ $E = 2\cos(3x) + 6\cos x - 4\cos^2 x - 4\cos x$ $E = 2\cos(3x) + 2\cos x - 4\cos^2 x$

Next, we use the double angle formula for cosine: $\cos(2x) = 2\cos^2 x - 1$. From this, $2\cos^2 x = 1 + \cos(2x)$. So, $4\cos^2 x = 2(1 + \cos(2x)) = 2 + 2\cos(2x)$.

Substitute this into the expression for $E$: $E = 2\cos(3x) + 2\cos x - (2 + 2\cos(2x))$ $E = 2\cos(3x) + 2\cos x - 2 - 2\cos(2x)$ $E = 2\cos(3x) - 2\cos(2x) + 2\cos x - 2$.

Now, we need to evaluate this expression with $x = \frac{\pi}{7}$. Consider the roots of the equation $z^7 - 1 = 0$. These are $z_k = e^{i \frac{2k\pi}{7}}$ for $k=0, 1, \dots, 6$. Since $z^7 - 1 = (z-1)(z^6 + z^5 + z^4 + z^3 + z^2 + z + 1)$, the roots of $z^6 + z^5 + z^4 + z^3 + z^2 + z + 1 = 0$ are $z_k = e^{i \frac{2k\pi}{7}}$ for $k=1, 2, \dots, 6$. Divide the equation $z^6 + z^5 + z^4 + z^3 + z^2 + z + 1 = 0$ by $z^3$ (since $z \neq 0$): $z^3 + z^2 + z + 1 + \frac{1}{z} + \frac{1}{z^2} + \frac{1}{z^3} = 0$ Group terms: $\left(z^3 + \frac{1}{z^3}\right) + \left(z^2 + \frac{1}{z^2}\right) + \left(z + \frac{1}{z}\right) + 1 = 0$. For $z = e^{i\theta}$, we know $z^n + \frac{1}{z^n} = 2\cos(n\theta)$. Let $\theta = \frac{2\pi}{7}$. Then $z = e^{i \frac{2\pi}{7}}$. Substituting into the grouped equation: $2\cos\left(3 \cdot \frac{2\pi}{7}\right) + 2\cos\left(2 \cdot \frac{2\pi}{7}\right) + 2\cos\left(\frac{2\pi}{7}\right) + 1 = 0$ $2\cos\left(\frac{6\pi}{7}\right) + 2\cos\left(\frac{4\pi}{7}\right) + 2\cos\left(\frac{2\pi}{7}\right) + 1 = 0$.

Now, relate these angles to $\frac{\pi}{7}$: $\cos\left(\frac{6\pi}{7}\right) = \cos\left(\pi - \frac{\pi}{7}\right) = -\cos\left(\frac{\pi}{7}\right)$. $\cos\left(\frac{4\pi}{7}\right) = \cos\left(\pi - \frac{3\pi}{7}\right) = -\cos\left(\frac{3\pi}{7}\right)$.

Substitute these back into the equation: $2\left(-\cos\frac{\pi}{7}\right) + 2\left(-\cos\frac{3\pi}{7}\right) + 2\cos\frac{2\pi}{7} + 1 = 0$ $-2\cos\frac{\pi}{7} - 2\cos\frac{3\pi}{7} + 2\cos\frac{2\pi}{7} + 1 = 0$. Rearranging this equation, we get: $2\cos\frac{2\pi}{7} - 2\cos\frac{3\pi}{7} - 2\cos\frac{\pi}{7} = -1$.

Let's compare this with our expression for $E$: $E = 2\cos\frac{3\pi}{7} - 2\cos\frac{2\pi}{7} + 2\cos\frac{\pi}{7} - 2$. Notice that the first three terms of $E$ are the negative of the left side of the equation we derived: $-(2\cos\frac{2\pi}{7} - 2\cos\frac{3\pi}{7} - 2\cos\frac{\pi}{7}) = -(-1) = 1$. So, $2\cos\frac{3\pi}{7} - 2\cos\frac{2\pi}{7} + 2\cos\frac{\pi}{7} = 1$.

Substitute this value back into the expression for $E$: $E = 1 - 2 = -1$.

Finally, we need to find the absolute value: $|E| = |-1| = 1$.