Find value of $(\cos\frac{2\pi}{7})^ {\frac{1}{3}} + (\cos\frac{4\pi}{7})^ {\frac{1}{3}} + (\cos\frac{8\pi}{7})^ {\frac{1}{3}} $

Question

The question of finding the value of $$(\cos\frac{2\pi}{7})^ {\frac{1}{3}} + (\cos\frac{4\pi}{7})^ {\frac{1}{3}} + (\cos\frac{8\pi}{7})^ {\frac{1}{3}} $$ was on my list. I was trying to apply the $n$-th roots of unity, but other ideas are welcome. I also tried Newton's sums, but it's not working.

I searched around here and I didn't find a similar one, but if they do, just say that I delete the topic.

Answer 1

This cubic-root sum, which has the inscrutable value

$$\left(\cos\frac{2\pi}{7}\right)^{\frac{1}{3}} +\left(\cos\frac{4\pi}{7}\right)^{\frac{1}{3}}+\left(\cos\frac{8\pi}{7}\right)^{\frac{1}{3}} = \left[ \frac{1}{2}\left(5-3\cdot 7^{\frac{1}{3}}\right) \right]^{\frac{1}{3}}$$ was known over a hundred years ago. Outlined below is an elementary evaluation of the sum.

Note that $\cos\frac{2\pi}7$, $\cos\frac{4\pi}7 $ and $\cos\frac{8\pi}7 $ are the roots of

$$x^3+\frac{1}{2}x^2-\frac{1}{2}x-\frac{1}{8}=0$$

and use the short-hands

$$c_1=\left(\cos\frac{2\pi}{7}\right)^{\frac{1}{3}},\space \space \space c_2=\left(\cos\frac{4\pi}{7}\right)^{\frac{1}{3}},\space \space \space c_3=\left(\cos\frac{8\pi}{7}\right)^{\frac{1}{3}} $$

to get

$$c_1^3+c_2^3+c_3^3 = -\frac{1}{2},\>\>\> c_1^3c_2^3+c_2^3c_3^3+c_3^3c_1^3 = -\frac{1}{2},\>\>\> c_1^3c_2^3c_3^3 = \frac{1}{8}\tag{1}$$

Then, let $$A=c_1+c_2+c_3, \space\space\space B=c_1c_2+c_2c_3+c_3c_1$$

and evaluate

$$A^3 = c_1^3+c_2^3+c_3^3 +3AB-3c_1c_2c_3,$$ $$B^3= c_1^3c_2^3+c_2^3c_3^3+c_3^3c_1^3+3c_1c_2c_3 AB-3(c_1c_2c_3)^2 $$ Plugging (1) into above expressions to get

$$A^3=-2+3AB,\space\space\space B^3=-\frac{5}{4}+\frac{3}{2}AB\tag{2}$$

Next, evaluate

$$(2AB-3)^3=8A^3B^3-36A^2B^2+54AB-27$$

and use the results (2) to get the equation satisfied by $A$,

$$\left( \frac{2A^3-5}{3}\right)^3 =-7\tag{3}$$

Finally, solve (3) and we obtain the sum,

$$A = \left[ \frac{1}{2}\left(5-3\cdot 7^{\frac{1}{3}}\right) \right]^{\frac{1}{3}}$$

Answer 2

If you begin with double your cosines, you have the three (real) roots of $$ x^3 + x^2 - 2x - 1 $$ so that the cosines themselves are the three roots of $$ 8 w^3 + 4 w^2 - 4 w - 1 $$

After that there is still a bunch of fiddling, since your numbers are the cube roots of these.

From 1875 book by Reuschle. The original technique is due to Gauss, before Galois. A modern treatment is in Galois Theory by Cox.