$$\frac{2\sqrt{7}}{3}\cos\left (\frac{1}{3}\arccos\left (\frac{1}{2\sqrt{7}} \right ) \right )+\frac{2}{3}$$
This is actually a solution for $x^3-2x^2-x+1$, and I tried to solve and got that solution.
And I have found this simplified solution $$1+2\cos\left(\frac{2\pi}{7}\right)$$
on the internet. However, I do not know how to even start to simplify it.
Any help what are the steps for that simplification?
Best Answer
Putting together the pieces discussed:
$\textbf{Step 1:}$
let $x = \frac13\arccos\frac{1}{2\sqrt{7}} \implies 4\cos^3 x - 3 \cos x = \frac{1}{\sqrt{28}}$
let $y = \frac{2\sqrt{7}}{3}\cos\left (\frac13\arccos\frac{1}{2\sqrt{7}}\right)+\frac23 \implies \frac{3y-2}{\sqrt{28}} = \cos x$
Substituting for $\cos x$ in first equation we get after simplification: $y^3 -2y^2 -y + 1 = 0$ .. Eqn (1)
$\textbf{Step 2:}$
Let $a = \cos \frac{2\pi}{7}, b = \cos \frac{\pi}{7}, c = \cos \frac{\pi}{14}$
We have $1 + 2\cos \frac{2\pi}{7} = 1 + 2a = 1 + 2(2b^2 - 1) = 4b^2 - 1$
Substituting this in Eqn (1) we get:
$(4b^2 - 1)^3 -2(4b^2 - 1)^2 - (4b^2 - 1) + 1 = 0$
Simplifying we get: $64b^6 - 80b^4 + 24b^2 -1 = 0$ ... Eqn(2)
$\textbf{Step 3:}$
$\cos 7t = \Re (\cos 7t + i\sin 7t) = \Re (\cos t + i\sin t)^7$
After simplifying and collecting real terms on right we get:
$\cos 7t = \cos t(64\cos^6 t - 112\cos^4 t + 56 \cos^2 t - 7)$ and substituting $t = \frac{\pi}{14}$ and noting $c = \cos \frac{\pi}{14}$ we get:
$64c^6 -112c^4 +56c^2 - 7 = 0$ ... Eqn(3)
We know $2c^2 = 1 + b$.
So we get $8(1+b)^3 - 28(1+b)^2 + 28(1+b) - 7 = 8b^3 -4b^2 - 4b + 1 = 0$
$\implies (8b^3 -4b)^2 = (4b^2-1)^2 \implies 64b^6 - 80b^4 + 24b^2 -1 = 0$
Thus we know $1+2\cos\frac{2\pi}{7}$ is a solution to Eqn (2)
$\textbf{Step 4:}$ We still have to prove that these two roots have the same value. For that we note that the derivative of the polynomial is given by $$3y^2-4y-1$$ which has zeros $\frac{2\pm\sqrt 7}{3}$. It follows that $y^3-2y^2-y+1$ is strictly increasing in $\left[\frac{2+\sqrt 7}{3},\infty\right)$ and hence it has at most one zero in this interval. In the previous steps we have seen that $\frac{2\sqrt{7}}{3}\cos\left (\frac13\arccos\frac{1}{2\sqrt{7}}\right)+\frac23$ and $1+2\cos\frac{2\pi}{7}$ both solve $y^3-2y^2-y+1=0$.
Moreover, we have $$0\leq \frac 13\arccos\frac{1}{2\sqrt 7}\leq\frac\pi 3$$ which implies $$\frac{2\sqrt{7}}{3}\cos\left (\frac13\arccos\frac{1}{2\sqrt{7}}\right)+\frac23\geq\frac{2\sqrt{7}}{3}\frac 12+\frac 23=\frac{2+\sqrt 7}{3}.$$ On the other hand we see that $\frac{2\pi}{7}<\frac{\pi}{3}$. This implies $$1+2\cos\frac{2\pi}{7}>1+2\cos\frac\pi3=2>\frac{2+\sqrt 7}{3}.$$ We can therefore conclude that they must indeed be equal.