Trigonometry Challenge 3

$\sec{\frac{4\pi}{9}}+\sec{\frac{2\pi}{9}}-\sec{\frac{\pi}{9}}=\frac{1}{\cos{\frac{4\pi}{9}}}+\frac{1}{\cos{\frac{2\pi}{9}}}-\frac{1}{\cos{\frac{\pi}{9}}}=\frac{\cos{\frac{2\pi}{9}}+\cos{\frac{4\pi}{9}}}{\cos{\frac{4\pi}{9}}\cos{\frac{2\pi}{9}}}-\frac{1}{\cos{\frac{\pi}{9}}}=\frac{2\cos{\frac{\pi}{3}}\cos{\frac{\pi}{9}}}{\cos{\frac{4\pi}{9}}\cos{\frac{2\pi}{9}}}-\frac{1}{\cos{\frac{\pi}{9}}}$

$=\frac{2(\frac{1}{2})\cos{\frac{\pi}{9}}}{\cos{\frac{4\pi}{9}}\cos{\frac{2\pi}{9}}}-\frac{1}{\cos{\frac{\pi}{9}}}=\frac{\cos{\frac{\pi}{9}}}{\cos{\frac{4\pi}{9}}\cos{\frac{2\pi}{9}}}-\frac{1}{\cos{\frac{\pi}{9}}}=\frac{\cos^{2}{\frac{\pi}{9}}-\cos{\frac{4\pi}{9}}\cos{\frac{2\pi}{9}}}{\cos{\frac{\pi}{9}}\cos{\frac{2\pi}{9}}\cos{\frac{4\pi}{9}}}=\frac{2\sin{\frac{\pi}{9}}\cos^{2}{\frac{\pi}{9}}-2\sin{\frac{\pi}{9}}\cos{\frac{4\pi}{9}}\cos{\frac{2\pi}{9}}}{2\sin{\frac{\pi}{9}}\cos{\frac{\pi}{9}}\cos{\frac{2\pi}{9}}\cos{\frac{4\pi}{9}}}$

$=\frac{\sin{\frac{2\pi}{9}}\cos{\frac{\pi}{9}}-\sin{\frac{\pi}{9}}(\cos{\frac{2\pi}{3}}+\cos{\frac{2\pi}{9}})}{\sin{\frac{2\pi}{9}}\cos{\frac{2\pi}{9}}\cos{\frac{4\pi}{9}}}=\frac{\sin{\frac{2\pi}{9}}\cos{\frac{\pi}{9}}-\sin{\frac{\pi}{9}}(-\frac{1}{2}+\cos{\frac{2\pi}{9}})}{\sin{\frac{2\pi}{9}}\cos{\frac{2\pi}{9}}\cos{\frac{4\pi}{9}}}=\frac{\sin{\frac{2\pi}{9}}\cos{\frac{\pi}{9}}-\sin{\frac{\pi}{9}}\cos{\frac{2\pi}{9}}+\frac{1}{2}\sin{\frac{\pi}{9}}}{\sin{\frac{2\pi}{9}}\cos{\frac{2\pi}{9}}\cos{\frac{4\pi}{9}}}$

$=\frac{\sin{(\frac{2\pi}{9}-\frac{\pi}{9})}+\frac{1}{2}\sin{\frac{\pi}{9}}}{\sin{\frac{2\pi}{9}}\cos{\frac{2\pi}{9}}\cos{\frac{4\pi}{9}}}=\frac{\sin{\frac{\pi}{9}}+\frac{1}{2}\sin{\frac{\pi}{9}}}{\sin{\frac{2\pi}{9}}\cos{\frac{2\pi}{9}}\cos{\frac{4\pi}{9}}}=\frac{2\sin{\frac{\pi}{9}}+\sin{\frac{\pi}{9}}}{2\sin{\frac{2\pi}{9}}\cos{\frac{2\pi}{9}}\cos{\frac{4\pi}{9}}}=\frac{(2+1)\sin{\frac{\pi}{9}}}{\sin{\frac{(2\times2)\pi}{9}}\cos{\frac{4\pi}{9}}}=\frac{3\sin{\frac{\pi}{9}}}{\sin{\frac{4\pi}{9}}\cos{\frac{4\pi}{9}}}$

$=\frac{2(3\sin{\frac{\pi}{9})}}{2\sin{\frac{4\pi}{9}}\cos{\frac{4\pi}{9}}}=\frac{(2\times 3)\sin{\frac{\pi}{9}}}{\sin{\frac{(2\times4)\pi}{9}}}=\frac{6\sin{\frac{\pi}{9}}}{\sin{\frac{8\pi}{9}}}=\frac{6\sin{\frac{\pi}{9}}}{\sin{(\pi-\frac{8\pi}{9})}}=\frac{6\sin{\frac{\pi}{9}}}{\sin{\frac{\pi}{9}}}=\boxed{6}$

$\begin{aligned} X & = \sec \frac {4\pi}9 + \sec \frac {2\pi}9 - \sec \frac \pi 9 \\ & = \frac 1{\cos \frac {4\pi}9} + \frac 1{\cos \frac {2\pi}9} - \frac 1{\cos \frac \pi 9} \\ & = \frac {\cos \frac {2\pi}9 \cos \frac \pi 9+\cos \frac {4\pi}9 \cos \frac \pi 9-\cos \frac {4\pi}9 \cos \frac {2\pi}9} {\cos \frac {4\pi}9 \cos \frac {2\pi}9 \cos \frac \pi 9} & \small \color{#3D99F6} \text{Note: } \frac {\cos(A-B)+\cos(A+B)}2 = \cos A \cos B \\ & = \frac {\cos \frac \pi 9 + {\color{#3D99F6}\cos \frac \pi 3}+\cos \frac \pi 3 + \cos \frac {5\pi}9 - \cos \frac {2\pi}9 - \cos \frac {2\pi}3}{2 \cos \frac \pi 9 \cos \frac {2\pi}9 \cos \frac {4\pi}9} \\ & = \frac {\cos \frac \pi 9 + {\color{#3D99F6}\cos \frac {3\pi}9} + \cos \frac {5\pi}9 {\color{#D61F06}- \cos \frac {2\pi}9} + 1}{2 \cos \frac \pi 9 \cos \frac {2\pi}9 \cos \frac {4\pi}9} & \small \color{#D61F06} \text{Note: } \cos (\pi - x) = - \cos x \\ & = \frac {\cos \frac \pi 9 +\cos \frac {3\pi}9 + \cos \frac {5\pi}9 {\color{#D61F06}+ \cos \frac {7\pi}9} + 1}{2 \color{#3D99F6} \cos \frac \pi 9 \cos \frac {2\pi}9 \cos \frac {4\pi}9} & \small \color{#3D99F6} \text{Note: }\prod_{k=0}^n \cos \left(\frac {2^k \pi}{2^{n+1}+1}\right) = \frac 1{2^{n+1}} \\ & = \frac {{\color{#D61F06}\cos \frac \pi 9 +\cos \frac {3\pi}9 + \cos \frac {5\pi}9 + \cos \frac {7\pi}9} + 1}{2 \color{#3D99F6} \times \frac 18} & \small \color{#D61F06} \text{Note: }\sum_{k=1}^n \cos \left(\frac {2k-1}{2n+1} \pi \right) = \frac 12 \\ & = \frac {\frac 32}{\frac 14} = \boxed{6} \end{aligned}$

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Note: $\displaystyle \sum_{k=0}^{n-1} \cos \left( \frac {2k+1}{2n+1}\pi \right) = \frac 12$ (see proof ).

Let $\Phi_9(x)=x^6-x^3+1$ be the 9th cyclotomic polynomial. We know that it has roots $x_k=e^{2 \pi k i / 9}$ for $k \in \{1,2,4,5,7,8\}$ . Then, we have: $x_k^6-x_k^3+1=0 \Longleftrightarrow x_k^3+\dfrac{1}{x_k^3}=1 \Longleftrightarrow \left(x_k+\dfrac{1}{x_k}\right)^3-3\left(x_k+\dfrac{1}{x_k}\right)-1=0$ . We also know that $c_k=x_k+\dfrac{1}{x_k}=2\cos\left(\dfrac{2 \pi k}{9}\right)$ , and since $\cos(\theta)=\cos(2\pi-\theta)$ , the polynomial $c^3-3c-1=0$ will have roots $c_k$ only for $k \in \{1,2,4\}$ .

The expression we want is $\dfrac{1}{\cos \frac{2\pi}{9}}+\dfrac{1}{\cos \frac{4\pi}{9}}+\dfrac{1}{\cos \frac{8\pi}{9}}$ , and by Vieta's formulas, that is $2\left(-\dfrac{-3}{1}\right)=\boxed{6}$ .