Question

$\cot^2 \frac{\pi}{7} + \cot^2 \frac{2\pi}{7} + \cot^2 \frac{3\pi}{7} = 5.$

Answer 1

5

Answer 2

To verify the identity $\cot^2 \frac{\pi}{7} + \cot^2 \frac{2\pi}{7} + \cot^2 \frac{3\pi}{7} = 5$, we can use the relationship between the roots of a polynomial and trigonometric functions.

1. Consider the equation $\sin(7x) = 0$.

   The solutions are $x = \frac{k\pi}{7}$ for any integer $k$. We can express $\sin(7x)$ in terms of $\sin x$ and $\cos x$ using De Moivre's theorem. $(\cos x + i \sin x)^7 = \cos(7x) + i \sin(7x)$ Expanding the left side using the binomial theorem and taking the imaginary part: $\sin(7x) = \binom{7}{1} \cos^6 x \sin x - \binom{7}{3} \cos^4 x \sin^3 x + \binom{7}{5} \cos^2 x \sin^5 x - \binom{7}{7} \sin^7 x$ Substitute the binomial coefficients: $\binom{7}{1} = 7$ $\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$ $\binom{7}{5} = \binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21$ $\binom{7}{7} = 1$ So, the equation becomes: $\sin(7x) = 7 \cos^6 x \sin x - 35 \cos^4 x \sin^3 x + 21 \cos^2 x \sin^5 x - \sin^7 x$

2. Form a polynomial in $\cot^2 x$.

   For $x = \frac{k\pi}{7}$ where $k \in \{1, 2, 3, 4, 5, 6\}$, we have $\sin(7x) = 0$ and $\sin x \neq 0$. Divide the equation by $\sin^7 x$: $0 = 7 \frac{\cos^6 x}{\sin^6 x} - 35 \frac{\cos^4 x}{\sin^4 x} + 21 \frac{\cos^2 x}{\sin^2 x} - 1$ Let $y = \cot^2 x$. The equation transforms into a cubic polynomial in $y$: $7y^3 - 35y^2 + 21y - 1 = 0$

3. Identify the roots of the polynomial.

   The roots of this cubic equation are the values of $\cot^2 x$ for $x = \frac{k\pi}{7}$, where $k \in \{1, 2, 3, 4, 5, 6\}$. However, due to the property $\cot^2(\pi - \theta) = (-\cot \theta)^2 = \cot^2 \theta$: $\cot^2(\frac{4\pi}{7}) = \cot^2(\pi - \frac{3\pi}{7}) = \cot^2(\frac{3\pi}{7})$ $\cot^2(\frac{5\pi}{7}) = \cot^2(\pi - \frac{2\pi}{7}) = \cot^2(\frac{2\pi}{7})$ $\cot^2(\frac{6\pi}{7}) = \cot^2(\pi - \frac{\pi}{7}) = \cot^2(\frac{\pi}{7})$ The angles $\frac{\pi}{7}, \frac{2\pi}{7}, \frac{3\pi}{7}$ are distinct and lie in the first quadrant, so their cotangent values are distinct and positive. Therefore, their squares $\cot^2(\frac{\pi}{7})$, $\cot^2(\frac{2\pi}{7})$, $\cot^2(\frac{3\pi}{7})$ are the three distinct roots of the cubic equation $7y^3 - 35y^2 + 21y - 1 = 0$.

4. Apply Vieta's formulas.

   For a cubic equation $ay^3 + by^2 + cy + d = 0$, the sum of the roots is given by $-b/a$. In our equation $7y^3 - 35y^2 + 21y - 1 = 0$: $a = 7$, $b = -35$, $c = 21$, $d = -1$. The sum of the roots is: $\cot^2 \frac{\pi}{7} + \cot^2 \frac{2\pi}{7} + \cot^2 \frac{3\pi}{7} = -\frac{-35}{7} = \frac{35}{7} = 5$.

Thus, the identity is verified.

Explanation of the solution:

The identity is proven by constructing a polynomial whose roots are the terms in the sum.

1. Start with the equation $\sin(7x) = 0$.
2. Express $\sin(7x)$ as a polynomial in $\sin x$ and $\cos x$ using De Moivre's theorem. $\sin(7x) = 7\cos^6 x \sin x - 35\cos^4 x \sin^3 x + 21\cos^2 x \sin^5 x - \sin^7 x$.
3. For $x = \frac{k\pi}{7}$ (where $k=1,2,3$), $\sin(7x)=0$ and $\sin x \neq 0$. Divide by $\sin^7 x$ to get an equation in $\cot^2 x$. $7\cot^6 x - 35\cot^4 x + 21\cot^2 x - 1 = 0$.
4. Let $y = \cot^2 x$. This yields a cubic equation: $7y^3 - 35y^2 + 21y - 1 = 0$.
5. The roots of this cubic equation are $\cot^2(\frac{\pi}{7})$, $\cot^2(\frac{2\pi}{7})$, and $\cot^2(\frac{3\pi}{7})$. (Note: $\cot^2(\frac{4\pi}{7}) = \cot^2(\frac{3\pi}{7})$, etc.)
6. By Vieta's formulas, the sum of the roots of $ay^3+by^2+cy+d=0$ is $-b/a$. Sum of roots $= -(-35)/7 = 5$.

Answer:

The given statement $\cot^2 \frac{\pi}{7} + \cot^2 \frac{2\pi}{7} + \cot^2 \frac{3\pi}{7} = 5$ is true.