Question

The value of $\frac{4}{2\cos^2\frac{\pi}{7}-\cos\frac{\pi}{7}-1}$ is

Answer 1

-16cos(π/7)

Answer 2

Let the given expression be $E$. $E = \frac{4}{2\cos^2\frac{\pi}{7}-\cos\frac{\pi}{7}-1}$

Let $x = \cos\frac{\pi}{7}$. The denominator is $2x^2 - x - 1$. We can factor the quadratic expression $2x^2 - x - 1$. The roots of $2x^2 - x - 1 = 0$ are $x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-1)}}{2(2)} = \frac{1 \pm \sqrt{1+8}}{4} = \frac{1 \pm 3}{4}$. The roots are $x_1 = \frac{1+3}{4} = 1$ and $x_2 = \frac{1-3}{4} = -\frac{1}{2}$. So, $2x^2 - x - 1 = 2(x-1)(x+\frac{1}{2}) = (x-1)(2x+1)$.

Substitute $x = \cos\frac{\pi}{7}$ back: The denominator is $(\cos\frac{\pi}{7}-1)(2\cos\frac{\pi}{7}+1)$. So the expression becomes: $E = \frac{4}{(\cos\frac{\pi}{7}-1)(2\cos\frac{\pi}{7}+1)}$

Now, let's establish a key identity related to $\cos\frac{\pi}{7}$. Consider the identity $\cos(4\theta) = -\cos(3\theta)$ for $\theta = \frac{\pi}{7}$. This is true because $\cos(4\frac{\pi}{7}) = \cos(\pi - 3\frac{\pi}{7}) = -\cos(3\frac{\pi}{7})$. Using the multiple angle formulas: $\cos(4\theta) = 2\cos^2(2\theta) - 1$ $\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$ Let $x = \cos\theta = \cos\frac{\pi}{7}$. Then $\cos(2\theta) = 2x^2 - 1$. So, $2(2x^2-1)^2 - 1 = -(4x^3-3x)$. $2(4x^4 - 4x^2 + 1) - 1 = -4x^3 + 3x$. $8x^4 - 8x^2 + 2 - 1 = -4x^3 + 3x$. $8x^4 - 8x^2 + 1 = -4x^3 + 3x$. Rearranging the terms, we get: $8x^4 + 4x^3 - 8x^2 - 3x + 1 = 0$.

We know that $x=\cos\frac{\pi}{7}$ is a root of this polynomial. The roots of $8x^4+4x^3-8x^2-3x+1 = 0$ are $\cos(\pi/7), \cos(3\pi/7), \cos(5\pi/7), \cos(7\pi/7) = -1$. So $(x+1)$ is a factor. $8x^4+4x^3-8x^2-3x+1 = (x+1)(8x^3-4x^2-4x+1) = 0$. Since $x=\cos\frac{\pi}{7} \neq -1$, it implies that $8x^3-4x^2-4x+1 = 0$. So, $8\cos^3\frac{\pi}{7}-4\cos^2\frac{\pi}{7}-4\cos\frac{\pi}{7}+1 = 0$.

Now, let's consider the denominator again: $D = (\cos\frac{\pi}{7}-1)(2\cos\frac{\pi}{7}+1)$. We need to simplify $D$. Let's use the identity $2\cos^2\theta-1 = \cos(2\theta)$. The denominator can be written as: $2\cos^2\frac{\pi}{7}-\cos\frac{\pi}{7}-1 = (2\cos^2\frac{\pi}{7}-1) - \cos\frac{\pi}{7} = \cos\frac{2\pi}{7} - \cos\frac{\pi}{7}$.

So the expression is: $E = \frac{4}{\cos\frac{2\pi}{7} - \cos\frac{\pi}{7}}$ We use the product-to-sum identity: $2\sin A\sin B = \cos(A-B) - \cos(A+B)$. Let $A = \frac{2\pi}{7}$ and $B = \frac{\pi}{7}$. Then $\cos\frac{2\pi}{7} - \cos\frac{\pi}{7} = -2\sin\left(\frac{\frac{2\pi}{7}+\frac{\pi}{7}}{2}\right)\sin\left(\frac{\frac{2\pi}{7}-\frac{\pi}{7}}{2}\right)$ $= -2\sin\left(\frac{3\pi}{14}\right)\sin\left(\frac{\pi}{14}\right)$.

So, $E = \frac{4}{-2\sin\frac{3\pi}{14}\sin\frac{\pi}{14}} = \frac{-2}{\sin\frac{3\pi}{14}\sin\frac{\pi}{14}}$. Using the product-to-sum identity again for the denominator: $\sin\frac{3\pi}{14}\sin\frac{\pi}{14} = \frac{1}{2}\left[\cos\left(\frac{3\pi}{14}-\frac{\pi}{14}\right) - \cos\left(\frac{3\pi}{14}+\frac{\pi}{14}\right)\right]$ $= \frac{1}{2}\left[\cos\frac{2\pi}{14} - \cos\frac{4\pi}{14}\right] = \frac{1}{2}\left[\cos\frac{\pi}{7} - \cos\frac{2\pi}{7}\right]$.

Substitute this back into $E$: $E = \frac{-2}{\frac{1}{2}\left[\cos\frac{\pi}{7} - \cos\frac{2\pi}{7}\right]} = \frac{-4}{\cos\frac{\pi}{7} - \cos\frac{2\pi}{7}} = \frac{4}{\cos\frac{2\pi}{7} - \cos\frac{\pi}{7}}$ This is the same expression. We need to find the value of $\cos\frac{2\pi}{7} - \cos\frac{\pi}{7}$.

Let's use the identity $8x^3-4x^2-4x+1=0$ where $x=\cos\frac{\pi}{7}$. We know that $\cos\frac{2\pi}{7} = 2\cos^2\frac{\pi}{7}-1 = 2x^2-1$. The expression in the denominator is $\cos\frac{2\pi}{7} - \cos\frac{\pi}{7} = (2x^2-1) - x = 2x^2-x-1$. This brings us back to the original denominator.

Let's use a different set of identities. Consider the sum $S = \cos\frac{\pi}{7} + \cos\frac{3\pi}{7} + \cos\frac{5\pi}{7}$. We know that $\cos\frac{5\pi}{7} = \cos(\pi - \frac{2\pi}{7}) = -\cos\frac{2\pi}{7}$. So $S = \cos\frac{\pi}{7} + \cos\frac{3\pi}{7} - \cos\frac{2\pi}{7}$.

Let's use the identity $1+2\sum_{k=1}^n \cos(2k\theta) = \frac{\sin((2n+1)\theta)}{\sin\theta}$. For $\theta=\frac{\pi}{7}$, $n=3$: $1+2(\cos\frac{2\pi}{7} + \cos\frac{4\pi}{7} + \cos\frac{6\pi}{7}) = \frac{\sin(\frac{7\pi}{7})}{\sin(\frac{\pi}{7})} = \frac{\sin\pi}{\sin(\frac{\pi}{7})} = 0$. $1 + 2\cos\frac{2\pi}{7} + 2\cos\frac{4\pi}{7} + 2\cos\frac{6\pi}{7} = 0$. We know $\cos\frac{4\pi}{7} = -\cos\frac{3\pi}{7}$ and $\cos\frac{6\pi}{7} = -\cos\frac{\pi}{7}$. So, $1 + 2\cos\frac{2\pi}{7} - 2\cos\frac{3\pi}{7} - 2\cos\frac{\pi}{7} = 0$. Divide by 2: $\frac{1}{2} + \cos\frac{2\pi}{7} - \cos\frac{3\pi}{7} - \cos\frac{\pi}{7} = 0$. This implies $\cos\frac{\pi}{7} + \cos\frac{3\pi}{7} - \cos\frac{2\pi}{7} = \frac{1}{2}$. This is the value of $S$ we were looking for.

Now, let's look at the denominator of our original expression $D = \cos\frac{2\pi}{7} - \cos\frac{\pi}{7}$. From the identity $1 + 2\cos\frac{2\pi}{7} - 2\cos\frac{3\pi}{7} - 2\cos\frac{\pi}{7} = 0$: $1 + 2(\cos\frac{2\pi}{7} - \cos\frac{\pi}{7}) - 2\cos\frac{3\pi}{7} = 0$. $1 + 2D - 2\cos\frac{3\pi}{7} = 0$. This means $2D = 2\cos\frac{3\pi}{7} - 1$. So $D = \cos\frac{3\pi}{7} - \frac{1}{2}$.

This doesn't seem to simplify to a simple integer.

Let's use the known value $\cos\frac{\pi}{7} + \cos\frac{2\pi}{7} + \cos\frac{3\pi}{7} = \frac{1}{2}$. We need to evaluate $\frac{4}{2\cos^2\frac{\pi}{7}-\cos\frac{\pi}{7}-1} = \frac{4}{\cos\frac{2\pi}{7} - \cos\frac{\pi}{7}}$.

From $\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}-\cos\frac{2\pi}{7} = \frac{1}{2}$, we can write: $\cos\frac{2\pi}{7} - \cos\frac{\pi}{7} = \cos\frac{3\pi}{7} - \frac{1}{2}$. So, $E = \frac{4}{\cos\frac{3\pi}{7} - \frac{1}{2}}$.

Let's test the identity $4\cos^3\frac{\pi}{7}-4\cos\frac{\pi}{7}+1 = 0$ derived earlier. Let's re-derive the polynomial for $\cos(\pi/7)$. From $1 + 2\cos\frac{2\pi}{7} + 2\cos\frac{4\pi}{7} + 2\cos\frac{6\pi}{7} = 0$. Let $x = \cos\frac{\pi}{7}$.

Let's use the identity $2\cos(A)\cos(B) = \cos(A+B) + \cos(A-B)$. Consider the denominator $D = \cos\frac{2\pi}{7} - \cos\frac{\pi}{7}$. We know $\cos(2A) - \cos(A) = -2\sin(3A/2)\sin(A/2)$. $D = -2\sin\frac{3\pi}{14}\sin\frac{\pi}{14}$. And the expression is $E = \frac{4}{-2\sin\frac{3\pi}{14}\sin\frac{\pi}{14}} = \frac{-2}{\sin\frac{3\pi}{14}\sin\frac{\pi}{14}}$. We also know $\sin\frac{3\pi}{14} = \cos(\frac{\pi}{2}-\frac{3\pi}{14}) = \cos\frac{4\pi}{14} = \cos\frac{2\pi}{7}$. And $\sin\frac{\pi}{14} = \cos(\frac{\pi}{2}-\frac{\pi}{14}) = \cos\frac{6\pi}{14} = \cos\frac{3\pi}{7}$. So, $E = \frac{-2}{\cos\frac{2\pi}{7}\cos\frac{3\pi}{7}}$.

We know $\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{3\pi}{7} = \frac{1}{8}$. So $\cos\frac{2\pi}{7}\cos\frac{3\pi}{7} = \frac{1}{8\cos\frac{\pi}{7}}$. $E = \frac{-2}{\frac{1}{8\cos\frac{\pi}{7}}} = -16\cos\frac{\pi}{7}$.

This is a specific value. Let's check if this is correct. The value of $\cos\frac{\pi}{7}$ is approximately $0.9$. So $-16 \times 0.9 = -14.4$. This is a numerical answer. The problem typically expects an integer or a simple rational number.

The most simplified form is $-16\cos(\pi/7)$.

Final answer is $\boxed{-16\cos\frac{\pi}{7}}$