# Evaluate this Trigonometric Expression: $\sqrt{\cos \frac{2\pi}{7}} + \sqrt{\cos \frac{4\pi}{7}} + \sqrt{\cos \frac{6\pi}{7}}$

Evaluate
$$\sqrt{\cos \frac{2\pi}{7}} + \sqrt{\cos \frac{4\pi}{7}} + \sqrt{\cos \frac{6\pi}{7}}$$

I found the following

• $\large{\cos \frac{2\pi}{7}+\cos \frac{4\pi}{7} + \cos \frac{6\pi}{7}=-\dfrac{1}{2}}$

• $\large{\cos \frac{2\pi}{7}\times\cos \frac{4\pi}{7} + \cos \frac{4\pi}{7}\times\cos \frac{6\pi}{7} + \cos \frac{6\pi}{7}\times\cos \frac{2\pi}{7}}=-\dfrac{1}{2}$

• $\large{\cos \frac{2\pi}{7}\times\cos \frac{4\pi}{7}\times\cos \frac{6\pi}{7}=\dfrac{1}{8}}$

Now, by Vieta's Formula's, $\large{\cos \frac{2\pi}{7}, \cos \frac{4\pi}{7} \: \text{&} \: \cos \frac{6\pi}{7}}$ are the roots of the cubic equation

$$8x^3+4x^2-4x-1=0$$

And, the problem reduces to finding the sum of cube roots of the solutions of this cubic.

For that, I thought about transforming this equation to another one whose zeroes are the cube roots of the zeroes of this cubic by making the substitution

$$x\mapsto x^3$$

and getting another equation

$$8x^9+4x^6-4x^3-1=0$$

However, this new equation will have some extra roots too and we can't directly use Vieta's to get the desired sum.

Also, it's given that the sum evaluates to a radical of the form

$$\sqrt{\frac{1}{d}(a-b\sqrt[b]{c})}$$

where $a, b, c \: \text{&} \: d \in \mathbb Z$

Thanks!

• There is an explicit formula for the roots of a cubic, so you might try that. But is there any reason to expect a clean formula? – Thomas Andrews Dec 12 '14 at 21:12
• @ThomasAndrews In the question, I have been given that the sum evaluates to $\sqrt{\frac{1}{d}( a - b\sqrt[b]{c})}$ where $a,b,c \: \text{&} \: d \in \mathbb Z$ – user196761 Dec 12 '14 at 21:16
• That seems like something to put in the question, then. Give us the information you have if you are seeking help. – Thomas Andrews Dec 12 '14 at 22:02
• @ThomasAndrews Sorry, I've added it now. – user196761 Dec 12 '14 at 23:02
• – Lucian Dec 12 '14 at 23:17

let $$x=\sqrt{\cos{\dfrac{2\pi}{7}}},y=\sqrt{\cos{\dfrac{4\pi}{7}}},z=\sqrt{\cos{\dfrac{6\pi}{7}}},$$ then we have $$\begin{cases} x^3+y^3+z^3=-\dfrac{1}{2}\\ (xy)^3+(yz)^3+(xz)^3=-\dfrac{1}{2}\\ (xyz)^3=\dfrac{1}{8} \end{cases}$$ use this identity $$a^3+b^3+c^3=(a+b+c)^3-3(a+b+c)(ab+bc+ac)+3abc$$ so $$\begin{cases} (x+y+z)^3-3(x+y+z)(xy+yz+xz)+3xyz=-\dfrac{1}{2}\\ (xy+yz+xz)^3-3(xy+yz+xz)[xyz(x+y+z)]+3x^2y^2z^2=-\dfrac{1}{2}\\ xyz=\dfrac{1}{2} \end{cases}$$ let $$u=x+y+z, v=xy+yz+xz$$ then we have $$\begin{cases} u^3-3uv+2=0\\ 4v^3-6uv+5=0 \end{cases}$$ so we have $$\Longrightarrow 4v^3-2u^3+1=0, v=\dfrac{u^3+2}{3u}$$ so $$4\left(\dfrac{u^3+2}{3u}\right)^3-2u^3+1=0\Longrightarrow 4u^9-30u^6+75u^3+32=0$$ let $t=u^3$,so we have $$4t^3-30t^2+75t+32=0$$ let $t=\dfrac{5}{2}-a$,then $$4\left(\dfrac{5}{2}-a\right)^3-30\left(\dfrac{5}{2}-a\right)^2+75\left(\dfrac{5}{2}-a\right)+32=0$$ $$\Longrightarrow 4a^3=\dfrac{189}{2}\Longrightarrow a=\dfrac{3\sqrt{7}}{2}$$ so $$u=x+y+z=\sqrt{\cos{\dfrac{2\pi}{7}}}+\sqrt{\cos{\dfrac{4\pi}{7}}}+\sqrt{\cos{\dfrac{6\pi}{7}}}=\sqrt{\dfrac{1}{2}\left(5-3\sqrt{7}\right)}$$
Try using this: $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ Let $a^3= \cos{\frac{2\pi}{7}},b^3= \cos{\frac{4\pi}{7}},c^3= \cos{\frac{6\pi}{7}}$.
• I'm sorry, but I still can't figure out what to do with the second factor in the $\text{R.H.S.}$ – user196761 Dec 12 '14 at 22:52