Subfield: Algebra


Question: Compute $\cos ^{3} \frac{2 \pi}{7}+\cos ^{3} \frac{4 \pi}{7}+\cos ^{3} \frac{8 \pi}{7}$.

Solution:
The identity $\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta$ can be rewritten into the power-reducing identity

$$
\cos ^{3} \theta=\frac{1}{4} \cos 3 \theta+\frac{3}{4} \cos \theta
$$



Thus if $D$ is the desired sum,

$$
\begin{aligned}
D & =\cos ^{3} \frac{2 \pi}{7}+\cos ^{3} \frac{4 \pi}{7}+\cos ^{3} \frac{8 \pi}{7} \\
& =\frac{1}{4}\left(\cos \frac{6 \pi}{7}+\cos \frac{12 \pi}{7}+\cos \frac{24 \pi}{7}\right)+\frac{3}{4}\left(\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{8 \pi}{7}\right) .
\end{aligned}
$$

Observe that $\cos \frac{24 \pi}{7}=\cos \frac{10 \pi}{7}$, so

$$
D=\frac{1}{4}\left(\cos \frac{6 \pi}{7}+\cos \frac{12 \pi}{7}+\cos \frac{10 \pi}{7}\right)+\frac{3}{4}\left(\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{8 \pi}{7}\right) .
$$

Notice also that $\cos \theta=\cos (2 \pi-\theta)$ implies $\cos \frac{12 \pi}{7}=\cos \frac{2 \pi}{7}, \cos \frac{10 \pi}{7}=\cos \frac{4 \pi}{7}$, and $\cos \frac{8 \pi}{7}=$ $\cos \frac{6 \pi}{7}$. Rewriting $D$ using the least positive equivalent angles yields

$$
\begin{aligned}
D & =\frac{1}{4}\left(\cos \frac{6 \pi}{7}+\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}\right)+\frac{3}{4}\left(\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}\right) \\
& =\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7} .
\end{aligned}
$$

To evaluate this sum, use the identity $\cos \theta=\cos (2 \pi-\theta)$ again to write

$$
2 D=\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}+\cos \frac{8 \pi}{7}+\cos \frac{10 \pi}{7}+\cos \frac{12 \pi}{7}
$$

If $\alpha=\cos \frac{2 \pi}{7}+i \sin \frac{2 \pi}{7}$, notice that the right side of the equation above is simply the real part of the sum $\alpha+\alpha^{2}+\alpha^{3}+\alpha^{4}+\alpha^{5}+\alpha^{6}$. Because $\alpha^{n}$ is a solution to the equation $z^{7}=1$ for $n=0,1, \ldots, 6$, the sum $1+\alpha+\alpha^{2}+\cdots+\alpha^{6}$ equals 0 . Hence $\alpha+\alpha^{2}+\cdots+\alpha^{6}=-1$ and $D=-1 / 2$.
Construct a cubic polynomial in $x$ for which $\cos \frac{2 \pi}{7}, \cos \frac{4 \pi}{7}$, and $\cos \frac{8 \pi}{7}$ are zeros; then the sum of their cubes can be found using techniques from the theory of equations. In particular, suppose the three cosines are zeros of $x^{3}+b x^{2}+c x+d$. Then

$$
\begin{aligned}
b & =-\left(\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{8 \pi}{7}\right) \\
c & =\cos \frac{2 \pi}{7} \cos \frac{4 \pi}{7}+\cos \frac{2 \pi}{7} \cos \frac{8 \pi}{7}+\cos \frac{4 \pi}{7} \cos \frac{8 \pi}{7}, \text { and } \\
d & =-\cos \frac{2 \pi}{7} \cos \frac{4 \pi}{7} \cos \frac{8 \pi}{7}
\end{aligned}
$$

Use complex seventh roots of unity (as in the previous solution) to find $b=1 / 2$. To find $c$, use the product-to-sum formula $2 \cos A \cos B=\cos (A+B)+\cos (A-B)$ three times:

$$
\begin{aligned}
2 c & =\left(\cos \frac{6 \pi}{7}+\cos \frac{2 \pi}{7}\right)+\left(\cos \frac{10 \pi}{7}+\cos \frac{6 \pi}{7}\right)+\left(\cos \frac{4 \pi}{7}+\cos \frac{12 \pi}{7}\right) \\
& \left.=\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}+\cos \frac{8 \pi}{7}+\cos \frac{10 \pi}{7}+\cos \frac{12 \pi}{7} \text { [because } \cos \theta=\cos (2 \pi-\theta)\right] \\
& =-1
\end{aligned}
$$



Thus $c=-1 / 2$.

To compute $d$, multiply both sides by $\sin \frac{2 \pi}{7}$ and use the identity $2 \sin \theta \cos \theta=\sin 2 \theta$ :

$$
\begin{aligned}
d \sin \frac{2 \pi}{7} & =-\sin \frac{2 \pi}{7} \cos \frac{2 \pi}{7} \cos \frac{4 \pi}{7} \cos \frac{8 \pi}{7} \\
& =-\frac{1}{2} \sin \frac{4 \pi}{7} \cos \frac{4 \pi}{7} \cos \frac{8 \pi}{7} \\
& =-\frac{1}{4} \sin \frac{8 \pi}{7} \cos \frac{8 \pi}{7} \\
& =-\frac{1}{8} \sin \frac{16 \pi}{7} .
\end{aligned}
$$

Because $\sin \frac{16 \pi}{7}=\sin \frac{2 \pi}{7}$, the factors on both sides cancel, leaving

$$
d=-1 / 8
$$

Thus $\cos \frac{2 \pi}{7}, \cos \frac{4 \pi}{7}$, and $\cos \frac{8 \pi}{7}$ are roots of $x^{3}+\frac{1}{2} x^{2}-\frac{1}{2} x-\frac{1}{8}$; so each value also satisfies the equation $x^{3}=-\frac{1}{2} x^{2}+\frac{1}{2} x+\frac{1}{8}$. Hence the desired sum can be rewritten as

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

Previous work has already established that $\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{8 \pi}{7}=-1 / 2$, so it remains to compute $\cos ^{2} \frac{2 \pi}{7}+\cos ^{2} \frac{4 \pi}{7}+\cos ^{2} \frac{8 \pi}{7}$. The identity $A^{2}+B^{2}+C^{2}=(A+B+C)^{2}-2(A B+B C+A C)$ allows the use of previous results: $\cos ^{2} \frac{2 \pi}{7}+\cos ^{2} \frac{4 \pi}{7}+\cos ^{2} \frac{8 \pi}{7}=(-1 / 2)^{2}-2(-1 / 2)=5 / 4$. Thus

$$
\cos ^{3} \frac{2 \pi}{7}+\cos ^{3} \frac{4 \pi}{7}+\cos ^{3} \frac{8 \pi}{7}=-\frac{1}{2}\left(\frac{5}{4}\right)+\frac{1}{2}\left(-\frac{1}{2}\right)+\frac{3}{8}=-\frac{1}{2} .
$$

Final answer: $-\frac{1}{2}$