Question
From the equation who's roots are $\cos\frac{\pi}9,\cos\frac{3\pi}9,\cos\frac{5\pi}9,\cos\frac{7\pi}9$ and hence prove
a) $8\cos\frac{\pi}9\cos\frac{5\pi}9\cos\frac{7\pi}9=1=8\cos\frac{\pi}9\cos\frac{2\pi}9\cos\frac{4\pi}9$
b) $\sec^4\frac{\pi}9+\sec^4\frac{2\pi}9+\sec^4\frac{4\pi}9=1104$
My Attempt:
Say $y=e^{i\pi(2k+1)/9}$
Thus $y^9+1=0$ has solutions $e^{i\pi(2k+1)/9}$ for $k\in\{0,1,\dots,8\}$
Let $y+1/y=2x\implies y=x\pm\sqrt{x^2-1}$
Thus $(x\pm\sqrt{x^2-1})^9+1=0$ has solutions $\cos\frac{(2k+1)\pi}{9}$ for $k\in\{0,1\dots8\}$
Thus $(x\pm\sqrt{x^2-1})^9+1=\prod_{k=0}^8(x-\cos\frac{(2k+1)\pi}{9})$
But if we put $x=0$, LHS is complex while RHS is Real
What am i doing wrong?
$$\implies x^4-3x^2+1=x(x^2-2)$$
$$\implies(x^4-3x^2+1)^2=x^2(x^2-2)^2$$
Set $\sec^2\dfrac{(2k+1)\pi}9=\dfrac4{x^2}=v$(say)
On replacement & simplification
$$ v^4-v^3(24+16)+v^2(144+32+64)+v(\cdots)+1=0$$
whose roots are $\sec^2\dfrac{(2k+1)\pi}9,k=0,1,2,3$
$$\sum_{k=1}^4\left(\sec^2\dfrac{(2k+1)\pi}9\right)^2$$
$$=\left(\sum_{k=1}^4\sec^2\dfrac{(2k+1)\pi}9\right)^2-2\sum_{k_1,k_2=0,1,2,3, k_1>k_2}\sec^2\dfrac{(2k_1+1)\pi}9\cdot\sec^2\dfrac{(2k_2+1)\pi}9$$
$$=40^2-2(144+32+64)=?$$
Now for $k=1,\sec^2\dfrac{(2k+1)\pi}9=?$
Finally $\sec(\pi-t)=-\sec t,\sec^2(\pi-t)=\sec^2t$
Set $t=\dfrac{2\pi}9,\dfrac{4\pi}9$