Question

The value of $\csc^2 \frac{2\pi}{7} + \csc^2 \frac{2\pi}{7} + \csc^2 \frac{4\pi}{7}$ is equal to 8

Answer 1

False

Answer 2

The question asks to verify if the statement "The value of $\csc^2 \frac{2\pi}{7} + \csc^2 \frac{2\pi}{7} + \csc^2 \frac{4\pi}{7}$ is equal to 8" is true or false. The expression can be simplified as $2 \csc^2 \frac{2\pi}{7} + \csc^2 \frac{4\pi}{7}$.

We use the identity $\csc^2 x = 1 + \cot^2 x$. So, the expression becomes: $2 \csc^2 \frac{2\pi}{7} + \csc^2 \frac{4\pi}{7} = 2 \left(1 + \cot^2 \frac{2\pi}{7}\right) + \left(1 + \cot^2 \frac{4\pi}{7}\right)$ $= 2 + 2 \cot^2 \frac{2\pi}{7} + 1 + \cot^2 \frac{4\pi}{7}$ $= 3 + 2 \cot^2 \frac{2\pi}{7} + \cot^2 \frac{4\pi}{7}$

We know that $\cot(\pi - x) = -\cot x$, which implies $\cot^2(\pi - x) = \cot^2 x$. Therefore, $\cot^2 \frac{4\pi}{7} = \cot^2 \left(\pi - \frac{3\pi}{7}\right) = \cot^2 \frac{3\pi}{7}$. Substituting this into the expression: $3 + 2 \cot^2 \frac{2\pi}{7} + \cot^2 \frac{3\pi}{7}$

Now, we use a standard identity for the sum of squares of cotangents: For an odd integer $n$, the sum of squares of cotangents is given by: $\sum_{k=1}^{(n-1)/2} \cot^2 \frac{k\pi}{n} = \frac{(n-1)(n-2)}{6}$ For $n=7$, $(n-1)/2 = (7-1)/2 = 3$. So, for $n=7$: $\cot^2 \frac{\pi}{7} + \cot^2 \frac{2\pi}{7} + \cot^2 \frac{3\pi}{7} = \frac{(7-1)(7-2)}{6} = \frac{6 \times 5}{6} = 5$

Let the given expression be $E$. We are checking if $E=8$. So, we need to check if $3 + 2 \cot^2 \frac{2\pi}{7} + \cot^2 \frac{3\pi}{7} = 8$. This implies $2 \cot^2 \frac{2\pi}{7} + \cot^2 \frac{3\pi}{7} = 5$.

From the identity, we know that $\cot^2 \frac{\pi}{7} + \cot^2 \frac{2\pi}{7} + \cot^2 \frac{3\pi}{7} = 5$. If the statement is true, then it must be that: $2 \cot^2 \frac{2\pi}{7} + \cot^2 \frac{3\pi}{7} = \cot^2 \frac{\pi}{7} + \cot^2 \frac{2\pi}{7} + \cot^2 \frac{3\pi}{7}$ Subtracting $\cot^2 \frac{2\pi}{7} + \cot^2 \frac{3\pi}{7}$ from both sides, we get: $\cot^2 \frac{2\pi}{7} = \cot^2 \frac{\pi}{7}$ This implies $\cot \frac{2\pi}{7} = \pm \cot \frac{\pi}{7}$. Since both $\frac{\pi}{7}$ and $\frac{2\pi}{7}$ lie in the interval $(0, \pi/2)$, their cotangent values are positive. Thus, we must have $\cot \frac{2\pi}{7} = \cot \frac{\pi}{7}$. However, the cotangent function is strictly decreasing in the interval $(0, \pi/2)$. Since $\frac{\pi}{7} < \frac{2\pi}{7}$, it must be that $\cot \frac{\pi}{7} > \cot \frac{2\pi}{7}$. Therefore, $\cot^2 \frac{\pi}{7} \neq \cot^2 \frac{2\pi}{7}$. This contradicts our assumption that the given statement is true.

Hence, the statement "The value of $\csc^2 \frac{2\pi}{7} + \csc^2 \frac{2\pi}{7} + \csc^2 \frac{4\pi}{7}$ is equal to 8" is false.

Note: If the question had been $\csc^2 \frac{\pi}{7} + \csc^2 \frac{2\pi}{7} + \csc^2 \frac{4\pi}{7}$, then the value would be: $(1 + \cot^2 \frac{\pi}{7}) + (1 + \cot^2 \frac{2\pi}{7}) + (1 + \cot^2 \frac{4\pi}{7})$ $= 3 + \cot^2 \frac{\pi}{7} + \cot^2 \frac{2\pi}{7} + \cot^2 \frac{3\pi}{7}$ $= 3 + 5 = 8$ This is a common identity, which suggests a possible typo in the original question. However, based on the question as stated, the value is not 8.