How does $\frac{2\sqrt{7}}{3}\cos\left (\frac13\arccos\frac{1}{2\sqrt{7}}\right)+\frac23$ simplify to $1+2\cos\frac{2\pi}{7}$?

Question

$$\frac{2\sqrt{7}}{3}\cos\left (\frac{1}{3}\arccos\left (\frac{1}{2\sqrt{7}} \right ) \right )+\frac{2}{3}$$ 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?

Answer 1

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.

Answer 2

So, the following proof is not complete but I'm posting it anyway, because it's too long for the comments and I hope someone will be able to fill in the missing part.

We check that $1+2\cos\left(\frac{2\pi}{7}\right)$ is a solution to $x^3-2x^2-x+1=0$. For that we compute $$\left(1+2\cos\left(\frac{2\pi}{7}\right)\right)^3-2\left(1+2\cos\left(\frac{2\pi}{7}\right)\right)^2-\left(1+2\cos\left(\frac{2\pi}{7}\right)\right)+1\\ =8\cos^3\left(\frac{2\pi}{7}\right)+4\cos^2\left(\frac{2\pi}{7}\right)-4\cos\left(\frac{2\pi}{7}\right)-1\\ =2\left(\cos\left(\frac{6\pi}{7}\right)+3\cos\left(\frac{2\pi}{7}\right)\right)+2\left(1+\cos\left(\frac{4\pi}{7}\right)\right)-4\cos\left(\frac{2\pi}{7}\right)-1\\ =2\cos\left(\frac{6\pi}{7}\right)+2\cos\left(\frac{4\pi}{7}\right)+2\cos\left(\frac{2\pi}{7}\right)+1=0.$$

To obtain the third line you can use the identities $\cos(3x)=4\cos^3(x)-3\cos(x)$ and $\cos(2x)=2\cos^2(x)-1$. Now, I don't actually know why the last part equals zero. According to this answer, it is a well known fact which is non-trivial. I haven't come up with a reliable source so far though.

Finally, you can argue that the polynomial has at most one zero in $\left[\frac{2+\sqrt{7}}{3},\infty\right)$. So, it suffices to show that both expressions are greater than $\frac{2+\sqrt{7}}{3}$.

Answer 3

The equation as given is a bit diffifult to work with. Things are less cumbersome if you define $x=y+1$, and then the cubic equation takes the form

$y^3+y^2-2y-1=0.$

With this form we show that the roots are twice the cosines of rational multiples of $\pi$.

Method 1

We plug in $y=2\cos\theta$ and multiply through by $\sin\theta$ to get

$8\cos^3\theta\sin\theta+4\cos^2\theta\sin\theta-4\cos\theta\sin\theta-\sin\theta=0.$ (Eq. 1)

And now apply the trigonometric sum-product relation iteratively:

$2\cos u\sin v=\sin(u+v)-\sin(u-v)$

So:

$2\cos\theta\sin\theta=\sin(2\theta)$

$4\cos^2\theta\sin\theta=2\cos\theta\sin(2\theta)=\sin(3\theta)+\sin\theta$

(Watch your signs: the coefficient of $\theta$ in $u-v$ is negative.)

$8\cos^3\theta\sin\theta=2\cos\theta(\sin(3\theta)+\sin\theta)=\sin(4\theta)+2\sin(2\theta)$

Plugging these into Eq. 1 gives

$[\sin(4\theta)+2\sin(2\theta)]+[\sin(3\theta)+\sin\theta]-2\sin(2\theta)-\sin\theta=0$

And then a miracle occurs: several terms cancel each other and all that remains is

$\sin(4\theta)+\sin(3\theta)=0.$

Thus $\sin(4\theta)=-\sin(3\theta)$ snd we conclude that $4\theta+3\theta$ must be a multiple of $2\pi$ or $4\theta-3\theta$ must be an odd multiple of $\pi$. Since $\theta=$ any multiple of $\pi$ implies $y=2\cos\theta=\pm2$ and this fails to satisfy the original cubic equation, the correct roots must be the remaining $\theta$ values divisible by $2\pi/7$ giving $y\in\{2\cos(2\pi/7),2\cos(4\pi/7),2\cos(6\pi/7)\}.$

Method 2

If one root of an odd-degree polynomial equation over the integers has the form $2\cos\theta$, Galois theory predicts that conjugate roots having the form $2\cos2^k\theta$ exist for as many values of $k$ as would be required to match the Galois symmetry of the equation.

Then if $y=2\cos\theta$ is indeed a root of our cubic equation, so should be $y^2-2=2\cos(2\theta)$.

So given

$y^3+y^2-2y-1=0$

we test whether

$(y^2-2)^3+(y^2-2)^2-2(y^2-2)-1=0.$

$y^6-5y^4+6y^2-1=0$

We have $y^3=-y^2+2y+1$, so $y^6=(-y^2+2y+1)^2=y^4-4y^3+2y^2+4y+1$ giving

$-4y^4-4y^3+8y^2+4y=-4y\color{blue}{(y^3+y-2y-1)}=0$

where the blue factor is zero by hypothesis and thus each root $y=2\cos\theta$ must be accompanied by a root $y^2-2=2\cos(2\theta)$.

So we have a sequence of roots

$y=2\cos\theta,y^2-2=2\cos(2\theta),y^4-4y^2+2=2\cos(4\theta),...$

where (with the cubic equation being irreducible), $y$ can't match $y^2-2$ and neither of these can match $y^4-4y^2+2=(y-1)(y^3+y^2-2y-1)+(-y^2-y+1)=-y^2-y+1$. So the argument doubling gives a cyclic sequence of the three roots and therefore $y=2\cos\theta$ is also $y=2\cos(8\theta)$.

By this method $\cos\theta=\cos(8\theta)$ and thus $(8\pm1)\theta$ must be a multiple of $2\pi$. But $\cos(2\pi/9)+\cos(4\pi/9)+\cos(8\pi/9)=0$ (the second term is also $\cos(-4\pi/9)$ and the arguments $-4\pi/9,2\pi/9,8\pi/9$ are equally spaced within the trigonometric period), and the roots should sum to $-1$. Therefore $y=2\cos\theta$ where $\theta$ is forced to be a multiple of $2\pi/7$.

Neither method specifies which multiple of $2\pi/7$ corresponds to the specific root value given in the question. That cannot be done by algebra or trigonometry alone. We must appeal to (the principal branch of) the inverse cosine being defined in such a way that

$-(1/3)+(2\sqrt7)/3\cos[\frac13\arccos(\frac1{2\sqrt7})]=2\cos(2\pi/7).$

Answer 4

Conjecture without proof.

Let $a,b,c,f,k,m\in \mathbb{Z}$. If for irreducible cubic $-f+ax+bx^2+cx^3=0$ exist integer solution of system $$ \begin{cases} a^2 + 3 b f = k\\ a b + 9 c f = k\\ b^2 - 3 a c = k \end{cases} $$ then exist factorization of cubic over extension $\cos(\dfrac{2\pi}{km})$ and $x$ is linearly dependent on $\cos(\dfrac{2\pi}{km})$.

Examples code for W.Mathematica:

Factor[1 - x - 2 x^2 + x^3, Extension -> Cos[2 Pi/(7)]]
Factor[1 - x - 2 x^2 + x^3, Extension -> Cos[2 Pi/(7*4)]]
Factor[-1 + x + 4 x^2 + x^3, Extension -> Cos[2 Pi/(13)]]
Factor[-1 + x + 4 x^2 + x^3, Extension -> Cos[2 Pi/(13*2)]]