# How to prove $1+\cos \left(\frac{2\pi}{7}\right)-4\cos^2 \left(\frac{2\pi}{7}\right)-8\cos^3 \left(\frac{2\pi}{7}\right) \neq 0$

The task is to prove the following non-equality by hand:

$$1+\cos \left(\frac{2\pi}{7}\right)-4\cos^2 \left(\frac{2\pi}{7}\right)-8\cos^3 \left(\frac{2\pi}{7}\right) \neq 0$$

Wolframalpha shows this, but I can't prove it.

• You don't need to prove sth you can just calculate it (as wolfram did) and see that it's false. – MorphhproM Mar 30 '16 at 17:22
• @MorphhproM Sure..Why not stop doing Math alltogether? – MathematicianByMistake Mar 30 '16 at 17:23
• @MorphhproM He's asking how to see it without a calculator. – S.C.B. Mar 30 '16 at 17:24
• There could be a typo. The "minimal polynomial" for $\cos(2\pi/7)$ is $1 + 4 x - 4 x^2 - 8 x^3$. So, if there were a $4$ on the $\cos(2\pi/7)$ term, you'd have an identity. – Blue Mar 30 '16 at 17:27
• @Mongol-genius: The typo might not be yours. Books make mistakes, too. – Blue Mar 30 '16 at 17:28

We prove a closely related result, which in particular shows there is a typo in the given equation.

The number $e^{2\pi i/7}$ is a root of $x^7=1$, and therefore of $x^6+x^5+\cdots+x+1=0$, or equivalently of $$(x^3+x^{-3})+(x^2+x^{-2})+(x+x^{-1})+1=0.\tag{1}$$ (We divided through by $x^3$.) Let $w=\frac{1}{2}(x+x^{-1})$.

Note that $x^3 +x^{-3}=8w^3-6w$ and $x^2+x^{-2}=4w^2-2$ and $x+x^{-1}=2w$ So our equation can be rewritten as $$8w^3+4w^2-4w-1=0.\tag{2}$$ Since $e^{2\pi i/7}$ is a root of (1), it follows that $\cos(2\pi/7)$ is a root of (2).

• As always whole different approach $+1$ – Archis Welankar Mar 30 '16 at 17:36
• A whole different approach to what, @ArchisWelankar(I thought they used this often)? Anyway, +1. – S.C.B. Mar 30 '16 at 17:39
• @K.Power That is not true in general. Notice the argument used shifts from talking about $x$ to talking about $w = \frac{x + x^{-1}}{2} = \frac{e^{it}+e^{-it}}{2} \equiv \cos(t)$. The last equation is in terms of $w$ not $x$. – Winther Mar 30 '16 at 17:59
• Let $\theta=2\pi/7$. Then $\cos\theta=\frac{1}{2}(e^{i\theta}+e^{-i\theta})$. I think this comes quite early, it is a direct consequence of the familiar identity $e^{i\theta}=\cos\theta+i\sin\theta$. Since $e^{i\theta}$ is a root of (1), and $w=\frac{1}{2}(x+x^{-1})$, it follows that $\cos\theta$ is a root of (2). – André Nicolas Mar 30 '16 at 18:00
• @Piquito: Compare the equation of my answer above with the equation of the question. If both were true, we would have $3\cos(2\pi/7)=0$. – André Nicolas Mar 30 '16 at 18:14

Here, we will prove the equation by @AndreNicolas and @Blue in a more elementary manner.

$$1+4\cos \left(\frac{2\pi}{7}\right)-4\cos^2 \left(\frac{2\pi}{7}\right)-8\cos^3 \left(\frac{2\pi}{7}\right) = 0$$

Note that since $\cos 3x=4\cos^3 x-3\cos x$, and since $\cos 2x=2\cos^2 x-1$, our equation simplifies to $$\cos \left(\frac{\pi}{7}\right) -\cos \left(\frac{2\pi}{7}\right) + \cos \left(\frac{3\pi}{7}\right) = \frac{1}{2}$$

Let $x=\frac{\pi}{7}$.

$$\cos x - \cos 2x + \cos 3x = \frac{1}{2}$$

$$\cos x + \cos 3x + \cos 5x = \frac{1}{2}$$ $$\cos x + \cos 3x + \cos 5x + ... = \frac{{\sin 2nx}}{{2\sin x}}$$ Since $n = 3$ $$\cos x + \cos 3x + \cos 5x = \frac{{\sin 6x}}{{2\sin x}}$$ $$\frac{{\sin 6x}}{{2\sin x}} = \frac{1}{2} \Leftrightarrow \sin 6x = \sin x$$ Which is true since $\sin (\pi-x)=\sin x$.

Or,similarly if $$K=\cos x - \cos 2x + \cos 3x$$ then $$K\sin\frac{\pi}{7}=\frac{\sin\frac{2\pi}{7}}{2}+\frac{\sin\frac{4\pi}{7}-\sin\frac{2\pi}{7}}{2}+\frac{\sin\frac{6\pi}{7}-\sin\frac{4\pi}{7}}{2}=\frac{\sin\frac{6\pi}{7}}{2}\implies K=\frac{1}{2}$$

Since $2\sin A\cos B=\sin(A+B)+\sin(A-B)$

• MXYMXY Please correct the identities cos(3x) and cos(2x) and your answer :) – Mongol-genius Mar 30 '16 at 21:17
• @Mongol-genius Does is work now? – S.C.B. Mar 30 '16 at 22:35
• MXYMXY --I think after you have changed the identities your answer must change: your equation simplifies to cos(2pi/7)+cos(4pi/7)+cos(6pi/7)= -1/2 – Mongol-genius Mar 30 '16 at 22:44
• MXYMXY - However Thanks for your essential solution :) – Mongol-genius Mar 30 '16 at 22:50
• @Mongol-genius That simplifies to the equation above, using $\cos (\pi -x)=-\cos x$. – S.C.B. Mar 30 '16 at 23:00

Putting $x=\cos(\frac{2\pi}{7})$ you have the polynomial $1+x-4x^2-8x^3\not= 0$ then $8x^3+4x^2-x-1\not=0$ from that $$8x^3+4x^2-x-1=4x^2(2x+1)-x-1-x+x=4x^2(2x+1)-(2x+1)+x=(2x+1)(4x^2-1)+x=(2x+1)^2(2x-1)+x$$ Since $(2x+1)^2>0$ and $x>\frac{1}{2}$ since $\frac{2\pi}{7}<\frac{\pi}{3}$ we have that $(2x-1)>0$ and a sum of positive numbers isn't zero.

I will address the problem as it has been stated not as to whether there is a conjectured or assumed typo.

First off the problem is equivalent to showing

$$\frac{(1+\zeta)}{\zeta^{2}(1+2\zeta)}\neq4$$ where $\zeta=cos(2\pi/7)$.

Since $1/2\lt\zeta\lt1$, the left hand side of the inequality is less than $3$ and therefore we are done.

Consider $p(x)=1+x-4x^2-8x^3$, that have only one real root, $x_0\approx0,4$. Now, calculating $x=\cos(2\pi/7)\approx0,6$ we find that $x\neq x_0$

• You are asked to prove the non-equality by hand. If you meant for these approximations to be computed by hand, then please indicate this together with (a sketch of) a method of finding this approximation by hand. – Wojowu Mar 30 '16 at 17:40
• Because is continuous, I apply Bolzano+ bisection – Martín Vacas Vignolo Mar 30 '16 at 17:42
• You also have to compute the cosine. Either way, this is supposed to be in your answer, not a comment. – Wojowu Mar 30 '16 at 17:43

We consider $$f(x) = x - 4x^2 - 8x^3$$

Notice that for $x \ge \frac{1}{2}, \>\> f(x) < -1$ siince $f(\frac{1}{2}) = -\frac{3}{2}$ and $f'(\frac{1}{2}) < 0$ and $f''(x) < 0$ for all $x > -1$

Now it's only important to show that $\cos \frac{2 \pi}{7} \ge \frac{1}{2}$

It should be relatively easy to show $\cos \frac{2 \pi}{7} \ge \cos \frac{\pi}{3}$

• Please Include a proof of why $x \ge \frac{1}{2}, \>\> f(x) < -1$, and In your answer, not a comment. – S.C.B. Mar 30 '16 at 17:44