Question

Reduce the following expression to a simplified rational: $\frac{1}{1 - \cos \frac{\pi}{9}} + \frac{1}{1 - \cos \frac{5\pi}{9}} + \frac{1}{1 - \cos \frac{7\pi}{9}}$

Answer 1

18

Answer 2

Let the given expression be $S$. We have: $S = \frac{1}{1 - \cos \frac{\pi}{9}} + \frac{1}{1 - \cos \frac{5\pi}{9}} + \frac{1}{1 - \cos \frac{7\pi}{9}}$

Consider the identity $\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$. Let $\theta = \frac{\pi}{9}$. Then $3\theta = \frac{\pi}{3}$, and $\cos(3\theta) = \cos(\frac{\pi}{3}) = \frac{1}{2}$. Let $x = \cos(\frac{\pi}{9})$. Substituting into the identity: $4x^3 - 3x = \frac{1}{2}$ $8x^3 - 6x = 1$ $8x^3 - 6x - 1 = 0$

The roots of this cubic equation are $x_1 = \cos(\frac{\pi}{9})$, $x_2 = \cos(\frac{5\pi}{9})$, and $x_3 = \cos(\frac{7\pi}{9})$.

The expression we need to simplify is $S = \frac{1}{1 - x_1} + \frac{1}{1 - x_2} + \frac{1}{1 - x_3}$. Let $y = 1 - x$. Then $x = 1 - y$. Substitute this into the cubic equation $8x^3 - 6x - 1 = 0$: $8(1-y)^3 - 6(1-y) - 1 = 0$ $8(1 - 3y + 3y^2 - y^3) - 6 + 6y - 1 = 0$ $8 - 24y + 24y^2 - 8y^3 - 7 + 6y = 0$ $-8y^3 + 24y^2 - 18y + 1 = 0$ Multiplying by -1, we get: $8y^3 - 24y^2 + 18y - 1 = 0$

The roots of this new cubic equation are $y_1 = 1 - x_1$, $y_2 = 1 - x_2$, and $y_3 = 1 - x_3$. We need to compute $S = \frac{1}{y_1} + \frac{1}{y_2} + \frac{1}{y_3}$.

Using Vieta's formulas for the cubic equation $Ay^3 + By^2 + Cy + D = 0$: Sum of products of roots taken two at a time: $y_1y_2 + y_2y_3 + y_3y_1 = \frac{C}{A}$ Product of roots: $y_1y_2y_3 = -\frac{D}{A}$

For $8y^3 - 24y^2 + 18y - 1 = 0$, we have $A=8$, $B=-24$, $C=18$, $D=-1$. $y_1y_2 + y_2y_3 + y_3y_1 = \frac{18}{8} = \frac{9}{4}$ $y_1y_2y_3 = -\frac{-1}{8} = \frac{1}{8}$

The sum $S$ can be written as: $S = \frac{y_2y_3 + y_1y_3 + y_1y_2}{y_1y_2y_3}$ $S = \frac{9/4}{1/8} = \frac{9}{4} \times 8 = 9 \times 2 = 18$.