A nice trignometric identity

Question

How to prove that:

$$\cos\dfrac{2\pi}{13}+\cos\dfrac{6\pi}{13}+\cos\dfrac{8\pi}{13}=\dfrac{\sqrt{13}-1}{4} $$

I have a solution but its quite lengthy, I would like to see some elegant solutions. Thanks!

Answer 1

As @Tunk-Fey said in the comment, we only need to show $4x^2 + 2x - 3 =0$ with $x = \cos\frac{2\pi}{13} + \cos\frac{6\pi}{13} + \cos\frac{8\pi}{13} $ since we know $x$ should be positive.

Take $\theta = \frac{2\pi}{13}$, $w = \exp(i\theta)$ and $$A = w + w^3 + w^4$$

$$B = w^{-1} + w^{-3} + w^{-4}$$ we have $A + B = 2x$, thus we only need to show $(A+B)^2 + A+B - 3 = 0$

We will need the following facts:

since we have $w^{13} - 1 = (w-1)\sum_{k=0}^{12}w^k=0$ thus $\sum_{k=0}^{12}w^k = 0$, dividing by $w^6$($\neq 0$) gives $$\sum_{k=1}^6(w^k + w^{-k}) + 1 =0$$ and we have also $$w^5 + w^{-5} = w^8 + w^{-8}$$ $$w^6 + w^{-6} = w^7 + w^{-7}$$

Simple computation gives \begin{align} (A+B)^2 =&w^2 + w^{-2} + w^{6} + w^{-6} + w^8 + w^{-8} + 6\\ &+2(w + w^{-1} + w^2 + w^{-2} + w^3 + w^{-3} + w^4 + w^{-4} + w^5 + w^{-5} + w^{7}+w^{-7}) \\ \end{align} replacing $w^{7}+w^{-7}$ by $w^{6}+w^{-6}$ gives \begin{align} (A+B)^2 = &w^2 + w^{-2} + w^{6} + w^{-6} + w^8 + w^{-8} + 6+2(-1) \\ =& w^2 + w^{-2} + w^5 + w^{-5} + w^6 + w^{-6} + 4 \end{align}

Now adding together $(A+B)^2$, $A+B$ and $-3$ allows to conclude.

Answer 2

Using complex numbers: $$\sum_{k=0}^{12}e^{2ik\pi/13}=0$$ So, $$1+e^{2i\pi/13}+e^{4i\pi/13}..e^{24i\pi/13}=0$$ For Real part, $$1+\cos(2\pi/13)+\cos(4\pi/13)...\cos(24\pi/13)=0$$ $$\cos(2\pi/13)+\cos(4\pi/13)...\cos(12\pi/13)=-1/2$$ $$\cos\dfrac{2\pi}{13}+\cos\dfrac{6\pi}{13}+\cos\dfrac{8\pi}{13}=-1/2-\cos\dfrac{4\pi}{13}+\cos\dfrac{10\pi}{13}+\cos\dfrac{12\pi}{13}$$ I think you can do it from here.

Answer 3

I shall finish it based on what Aditya did, and all the credit goes to him. If you simply write $a_{2k}$ to denote $\cos(2k\pi/13)$, the identity you have is: $a_0+a_2+a_4+\dots+a_{24}=0$. You also need the basic identities: $$a_{2k}=a_{(26-2k)},\quad a_{2k}=2a_{k}^2-1,\quad 2a_{2k}a_{2j}=a_{2(k+j)}+a_{2(k-j)}$$

Now $$\begin{split} 0&=a_0+a_2+a_4+\dots+a_{24}\\ &=a_0+(a_2+a_6+a_8)+(a_{24}+a_{20}+a_{18})+\\ &(a_4+a_{12}+a_{16})+(a_{22}+a_{14}+a_{10})\\ &=1+2(a_2+a_6+a_8)+2(a_4+a_{12}+a_{16}) \end{split} $$ What you want is $x:=a_2+a_6+a_8$. Lets denote $y:=a_4+a_{12}+a_{16}$. So you have: $$1+2x+2y=0$$ But we also have: $$x^2=\frac{3}{2}+x+\frac{3}{2}y$$ That is, your $4x^2+2x-3=0$

Answer 4

I add another answer although it is not different in principle from some of the others. We have $$t=\cos \frac{2\pi}{13}+\cos \frac{6\pi}{13}+\cos \frac{8\pi}{13}$$ it is natural to then consider the other even divisions, $$s=\cos \frac{4\pi}{13}+\cos \frac{10\pi}{13}+\cos \frac{12\pi}{13}$$

Now $$t+s=\cos \frac{2\pi}{13}+\cos \frac{4\pi}{13}+\cos \frac{6\pi}{13}+\cos \frac{8\pi}{13}+\cos \frac{10\pi}{13}+\cos \frac{12\pi}{13}=\frac{\cos\frac{7\pi}{13}\sin\frac{6\pi}{13}}{\sin\frac{\pi}{13}}$$

$$\frac{\cos\frac{7\pi}{13}\sin\frac{6\pi}{13}}{\sin\frac{\pi}{13}}=\frac{1}{2}\frac{\sin\frac{13\pi}{13}-\sin \frac{\pi}{13}}{\sin\frac{\pi}{13}}=-\frac{1}{2}$$

Next we calculate $st$ using $\cos A \cos B=\frac{1}{2}(\cos (A+B)+\cos(A-B))$ we find $st=\frac{3}{2}(s+t)=-\frac{3}{4}$

So $s$ and $t$ are solutions to $4x^2+2x-3=0$ whose roots are $\frac{\sqrt{13}-1}{4}$ and $\frac{-\sqrt{13}-1}{4}$

Note that $$t=\cos \frac{2\pi}{13}+\cos \frac{6\pi}{13}-\cos \frac{5\pi}{13}$$ and since $\cos \frac{5\pi}{13}<\cos \frac{6\pi}{13}$ we have that $t>0$ and thus

$$\cos \frac{2\pi}{13}+\cos \frac{6\pi}{13}+\cos \frac{8\pi}{13}=\frac{\sqrt{13}-1}{4}$$

Answer 5

Consider the polynomial $P=X^{13}-1$, and discuss their roots